Chemistry 12 UNIT V ELECTROCHEMISTRY v. 1.1

Chemistry 12 UNIT V ELECTROCHEMISTRY v. 1.1 I. INTRODUCTION Put a strip of copper into a concentrated solution of nitric acid and it will quickly begin to bubble, turning the solution green and eating away the metal. Put an identical strip of copper into a concentrated solution of hydrochloric acid, and nothing happens. The reason is based on ELECTROCHEMISTRY ELECTROCHEMISTRY Branch of chemistry concerned with the conversion of chemical energy to electrical energy (and vice versa) ELECTROCHEMICAL REACTIONS Reactions that involve the transfer (loss and gain) of electrons between substances. Often setup as a cell a system that requires or produces electrical energy Consider the following reaction 2 Al(s) + 3 CuCl 2 (aq) 2AlCl 3 (aq) + 3 Cu(s) ionic: 2 Al(s) + 3 Cu 2+ (aq) + 2Cl - (aq) 2Al 3+ (aq) + 2Cl - (aq) + 3 Cu(s) net: 2 Al(s) + 3 Cu 2+ (aq) 2Al 3+ (aq) + 3 Cu(s) This reaction can be re-written as two separate HALF-REACTIONS. OXIDATION A half-reaction in which a species LOSES ELECTRONS OXIDATION HALF-REACTION Al(s) Al 3+ (aq) + 3e- REDUCTION A half-reaction in which a species GAINS ELECTRONS REDUCTION HALF-REACTION Cu 2+ (aq) + 2e- Cu(s) LEO the lion says GER LEO - Loss of Electrons is Oxidation GER - Gain of Electrons is Reduction

When a substance becomes OXIDIZED: It LOSES electrons (becoming more positive/less negative) and donates those electrons to another species, so that other speicies becomes reduced. An oxidation causes a reduction. -2- A substance being oxidized is referred to as the REDUCING AGENT. Ex. Zn Zn 2+ + 2e- U 3+ U 4+ + e- 2Cl - Cl 2 + 2e- When a substance becomes REDUCED: It GAINS electrons (becoming more negative/less positive) and accepts electrons from another species which is being oxidized. A reduction causes an oxidation. A substance being reduced is referred to as the OXIDIZING AGENT Ex. Cu 2+ +2e- Cu V 3+ + e- V 2+ F 2 + 2e- 2F - For every reduction, there must be an accompanying oxidation. A substance can t accept electrons if another substance won t give them off in the first place. REDUCTION-OXIDATION REACTION Consists of 2 half-reactions. One reduction and one oxidation. Called a REDOX REACTION. Ex. Consider the following reaction: 2 Al + 3 Cu 2+ 2 Al 3+ + 3 Cu Al: Al Al 3+ + 3e- Al is being oxidized, causing Cu 2+ to be reduced. Al is the REDUCTION AGENT Cu 2+ : Cu 2+ + 2e- Cu Cu 2+ is being reduced, causing Al to be oxidized. Cu 2+ is the OXIDIZING AGENT THE REDUCING AGENT IS OXIDIZED IN A REACTION THE OXIDIZING AGENT IS REDUCED IN A REACTION ANY TIME THAT YOU SEE THAT AN ATOM OR ION HAS CHANGED ITS CHARGE DURING A REACTION, YOU ARE DEALING WITH A REDOX REACTION.

Ex. Are the following atoms being oxidized or reduced (and would that make them the oxidizing agent or the reducing agent?) -3- a. I - I 3+ + 4eb. Au 3+ + 3e- Au c. Cu + Cu 2+ + e- d. F 2 + 2e- 2 F- Ex. For the following reactions indicate which substance is: a. being oxidized, b. being reduced, c. the oxidizing agent, d. the reducing agent. Zn 2+ + Mg Zn + Mg 2+ 2 Na + Br 2 2 Na + + 2 Br - II. OXIDATION NUMBERS OXIDATION NUMBER The charge that an atom would possess if the species containing the atom were made up of ions. (Recall combining capacity) You will need to be able to calculate the oxidation numbers of various atoms. RULES FOR DETERMINING OXIDATION NUMBER 1. The sum of the positive charges and the negative charges must equal the overall charge on the species. (In elemental form = charge is zero/neutral) 2. Oxidation number is related to an atoms position on the periodic table (has to do with the number of e- available for bonding) i. Group I ions (Li, Na, K Alkali metals) ALWAYS have an oxidation of 1+ when combined with other elements.

-4- ii. Group II ions (Ca, Ba, Mg Alkaline Earth Metals) ALWAYS have an oxidation number of 2+ when bonded in a mlc. iii. Group XVII (Cl, Br, F Halogens) TEND to have an oxidation number of 1- when bonded in a compound. (Many exceptions) 3. The oxidation number of oxygen is USUALLY 2-. (The exception is in peroxides (H 2 O 2 ) where the oxidation number is 1-.) 4. The oxidation number of hydrogen is USUALLY 1+. (Exception is metal hydrides (NaH, BaH 2 ) where the oxidation number is 1-) 5. All other oxidation numbers are variable. Each particular element has certain tendencies and possibilities according to their position on the periodic table, but should NEVER be assumed. 6. ELEMENTS IN THEIR ELEMENTAL FORM (NOT COMBINED WITH ANYTHING) ALWAYS HAVE AN OXIDATION NUMBER OF 0 UNLESS OTHERWISE STATED. Ex. Determine the oxidation number of each atom for the following: a. Na c. N 2 O 4 b. CCl 4 d. PO 4 3- Try: Determine the oxidation number of each atom for the following: a. P 4 c. H 4 P 2 O 7 b. PbSO 4 d. NH 4 + An atoms change in oxidation number indicates if it is being reduced or oxidized. When it becomes more positive, it is being oxidized When it becomes more negative, it is being reduced

Consider the following -5- ClO 3 - ClO 4 - H 2 O 2 H 2 O Cr 3+ CrO 4 2- (Cl 5+ Cl 7+ = oxidation) NO 2 N 2 O 3 TIP if the ratio of attached oxygen atoms increases in a reaction oxidation If the ratio of attached oxygen atoms decreases in a reaction reduction III. PREDICTING THE SPONTANEITY OF A REDOX REACTION There is a table called the Table of Standard Reduction Potentials, which is partially shown below. F 2 + 2 e- 2 F - Ag + + e- Ag Cu 2+ + 2 e- Cu Zn 2+ + 2 e- Zn Li + + e- Li The following observations can be made from the Table of Standard Reduction Potentials. (These will help you locate a half-reaction much more quickly) 1. In general, metals are found in the bottom half of the table on the right (Exceptions Cu, Ag, Hg, Au are all clustered together on the table) 2. In general, the halogens and oxyanions (anions containing oxygen) are found in the top half of the table on the right. 3. Some metals (Fe, Sn, Cr, Hg, Cu) have more than one common oxidation number. In these cases there will be MORE THAN ONE POSSIBLE HALF REACTION shown on the table involving these metals. A given ion may be found on either side of the table.

Ex. Copper Cu + + e- Cu(s) Cu 2+ + 2e- Cu(s) Cu 2+ + e- Cu + -6-4. H 2 O 2 is found at the top left side (+1.78 V) and lower down on the right side (+0.70 V) Watch for exceptions. Oxidizing agents are found on the left of the table. Reducing agents are found on the right. ALL reactions on the table are written as REDUCTIONS ( ) they gain e- in the forward rxn The Table of Standard Reduction Potentials is used in a similar way to the Table of Relative Strengths of Acids and Bases. The speices in the upper left corner have a tendency to proceed in the forward direction, while the species in the lower right corner have a tendency to proceed in ther reverse direction. Each of the reactions on the standard reduction table could proceed in either direction, providing the correct substances are present. Ex. The half reaction for Zn and Zn 2+ is: Zn 2+ + 2e- Zn If there was a piece of Zn(s) in a solution of Zn 2+ (aq) you can write either: Zn 2+ + 2e- Zn (written as a reduction) Or Zn Zn 2+ + 2e- (written as an oxidation) Note When referring to an ISOLATED half-reaction, use equilibrium arrows to show that the reaction can go forward or backwards. Ag + + e- Ag If the half-reaction is made to UNDERGO EITHER REDUCTION OR OXIDATION AS A RESULT OF BEING PART OF A REDOX REACTION, then use a one-way reaction arrow.

Try: State if the following ions could undergo reduction, oxidation, or both: -7- a. Ni 2+ c. NO e. Fe 2+ b. Cl - d. Cu + f. Al 3+ Become familiar with the idea that reduction equations are those on the table in the forward direction and oxidation equations are those in the reverse direction. Assume you have two different half-reactions in two different beakers. In one beaker there is Zn(s) in a solution of Zn 2+, and in the other is Cu(s) in a solution of Cu 2+. The two possible half-reactions are: Zn 2+ + 2e- Zn Cu 2+ + 2e- Cu Of the two oxidizing agents (Cu 2+ and Zn 2+ ), the Cu 2+ is higher on the left side of the table, and therefore has a greater tendency to become reduced. The reduction reaction will be: Cu 2+ + 2e- Cu Of the two reducing agents ( Cu and Zn ), the Zn is lower on the right side of the table, and therefore has a greater tendency to be oxidized. The oxidation reaction will be: Zn 2+ + 2e- Zn or rewritten as Zn Zn 2+ + 2e- Recall: Reduction and oxidation must BOTH occur for a redox reaction to happen. Therefore the two half-reactions must be connected or joined to allow the transfer of electrons to occur (more later) The overall reaction, which in this particular case will occur spontaneously, is found by adding together the two half reactions ONE OXIDATION AND ONE REDUCTION. Cu 2+ + 2e- Cu add the two ½ rexns (written correctly) Zn Zn 2+ + 2e- cross out what appears the same on both sides

-8- When two complete half-reactions (having all species present for each half-reaction) the higher half reaction on the table will undergo reduction and the lower will undergo oxidation. (Tip look for the strong OA and the stronger RA present) Try: Which is the stronger oxidizing agent in each of the following pairs: a. Ag + or Cu 2+ b. Co 2+ or Au 3+ Try: Which is the stronger reducing agent in each of the following pairs: a. H 2 O or H 2 O 2 b. Sn 2+ or Cu + Ex. Assuming you have the following species present: Br 2, Br -, I 2, I -, take the following half reactions and determine the overall redox reaction: Br 2 + 2e - 2Br - I 2 + 2e - 2I - If you are only given two species (or ions) rather than the four needed for two complete half-reactions as given above, you will need to be able to predict if that particular redox reaction will occur. RULES FOR PREDICTING SPONTANEOUS REDOX REACTIONS 1. If only one species in a half reaction is present, don t assume that the other species is also present. You need to be explicitly told if the other species is present. Ex. You are told a solution contains Cl -. Referring to the table, you see the half reaction Cl 2 +2e- 2Cl - Just because it is on the table doesn t mean it will happen. You have to have the reactants present in order for a reaction to occur. 2. If you are only given two potential reactants, rather than complete half-reactions, a reaction may or may not occur. In order to determine if a reaction will occur, the first thing to do is: Locate each reactant on the table (on any side). Take note of the position which is higher or lower (stronger RA / OA)

There are three possibilities: If BOTH species are OXIDIZING AGENTS or BOTH species are REDUCING A. AGENTS, then NO REACTION OCCURS. (Must have one reduction and one oxidation) -9- Ex. Assume the only reactants are Zn and Cu. Both are only found as RA. No rxn can occur Ex. Assume the only reactants are Br - and Cl -. B. If one is on the left of the table and one is on the right of the table two different cases are possible: i. The oxidizing agent (LEFT column) is HIGHER than the reducing agent (RIGHT column) THE REACTION WILL BE SPONTANEOUS Ex. Assume the reactants in a vessel are Cu 2+ and Zn. Cu 2+ is an OA, and is higher than Zn which is the RA. The reaction will spontaneously occur. ii. The oxidizing agent (LEFT column) is LOWER than the reducing agent (RIGHT column) THE REACTION WILL NOT OCCUR SPONTANEOUSLY (it is possible, but not spontaneous) Ex. Assume the reactants are Zn 2+ and Cu. SUMMARY A REACTION WILL ONLY BE SPONTANEOUS IF THE OXIDIZING AGENT IS ABOVE THE REDUCING AGENT.

Try: Which of the following species will oxidize Cu +? -10- Ni 2+ Ag + Pb 2+ I 2 H 2 O Try: Predict whether or not a reaction will occur when the following are mixed: a. Cl 2 and Br - b. Sn and Mn c. Ni 2+ and Pb d. Cl 2 and K COMMENT ON H + Some half-reactions require H + to occur. Ex. H 2 O 2 + 2H + +2e- 2H 2 O If H + is present in a particular half-reaction, it must be treated like any other reactant. For example, if you are asked whether the SO 4 2- in a solution of Na 2 SO 4 will reduce: SO 4 2- + 4H+ + 2e- H 2 SO 4 + 2H 2 O Your answer should be there is no reaction unless H + is could is present also (or likewise only if the solution is acidic ), just as nothing would happen if SO 4 2- wasn t present same with H +. There will be no reaction if it were not present. Note that H + is necessary in many reduction half-reaction, but there is also one half-reaction where it is the only substance involved. 2H + + 2e- H 2

IV. BALANCING HALF-REACTIONS -11- A half-reaction must be balanced, just as other chemical reactions must be balanced, however halfreactions are balanced for mass and CHARGE. Balancing half-reactions is not overly complicated, but it is very easy to make mistakes if you are careless about writing the charges on ions. BALANCING HALF-REACTION STEPS Usually when you are required to balance a half-reaction, you will be given a skeleton equation containing the major atoms. It is up to you to complete the balancing by applying other species as follows: 1. Balance the MAJOR ATOMS (atoms other than Oxygen or Hydrogen) 2. Balance the OXYGEN ATOMS by ADDING WATER MOLECULES Most redox rxns occus in aqueous sol n. Any oxygen that is used as a reactant or given off as a product will be in the form of water, NOT O or O 2. 3. Balance the HYDROGEN ATOMS BY ADDING H +. Redox reaction are (initially) treated as if they occur in acidic conditions. (Basic will require one extra step later) Since water contains hydrogen, always balance the hydrogen after you have added water to balance the oxygen. 4. Balance the OVERALL CHARGE BY ADDING ELECTRONS. The total charge (+ve, -ve, or neutral) must be the same on both sides. Do this after the hydrogens, as they have a charge. Note NEVER vary the order of balancing, it will make it difficult to impossible if you don t follow the above order EXACTLY. Memory aid: MAJOR Major atoms HYDROXIDE O H - (charge)

Ex. Balance the half-reaction: RuO 2 Ru -12- Ex. Balance the half-reaction: Cr 2 O 7 2- Cr 3+ Try: Balance the half-reaction: MnO 4 - Mn 2+ BALANCING BASIC HALF-REACTIONS All the above solutions were assumed to be in acidic solutions. Sometimes you will be required to balance half-reactions in BASIC conditions. First, balance as if it were in acidic conditions Ex. Pb HPbO 2-1. Balance the major atoms 2. Balance the oxygen atoms 3. Balance the hydrogen atoms 4. Balance the charge.

Now, CONVERT THE EQUATION TO BASIC CONDITIONS which is done by: -13-5. Adding the water equilibrium equation in such a way as to CANCEL OUT ALL THE H + in the half reaction. H + + OH - H 2 O or H 2 O H + + OH - Try: Balance the following: a. HC 2 H 3 O 2 C 2 H 5 OH in acidic conditions b. MnO 4 - MnO 2 in basic conditions V. BALANCING REDOX EQUATIONS USING HALF-REACTIONS Note: There are two common methods for balancing redox equations: using half reaction and using oxidation numbers. You are not required to know both methods (as they end up with the same results) but it may be to your advantage to be familiar with both of them. Balancing using half-reactions is easier for more complicated redox equations, the oxidation number is easier for simpler redox equations. STEPS IN BALANCING EQUATIONS USING HALF REACTIONS Ex. Balance ClO 4 - + I 2 Cl - + IO 3 - in acidic solution. 1. Break into the two half reactions (look for similar species)

-14-2. Balance the individual half reactions separately. 3. Multiply the half reactions by whole number so as to make the TOTAL ELECTRONS LOST = TOTAL ELECTRONS GAINED (same number for both) 4. Add the two half reactions, canceling out any species common to both sides Work THOUROUGHLY AND CAREFULLY. One small mistake can throw the whole question off. Double check at the end. (atoms/e-/charrges balanced?) Try: Balance MnO 4 - + C 2 O 4 2- MnO 2 + CO 2 in basic solution.

-15- DISPROPORTIONATION REACTION A redox reaction in which the SAME SPECIES IS BOTH REDUCED AND OXIDIZED. Ex. Balance ClO 2 - ClO 3 - + Cl - in basic solution. STEPS IN BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS This method is somewhat of a shortcut, based on that fact that since the total number of electrons lost in an oxidation half-reaction must equal the total number of electrons gained in a reduction halfreaction, so the following two statements are true: An increase in oxidation number in one species must be balanced by a decrease in oxidation number of a second species. In any redox reaction, the overall change in oxidation number must equal zero. Ex. Balance the following redox reaction ClO 4 - + I 2 Cl - + IO 3-1. Balance the MAJOR ATOMS 2. Assign OXIDATION NUMBER TO ALL STOMS INVOLVED IN A CHANGE (Δ ON) and determine the TOTAL CHANGE (this includes the number of atoms that changed also)

-16-3. Balance the change in oxidation number by multiplying to get the lowest common multiple for each half reaction. 4. Balance OXYGEN ATOMS by adding H 2 O and HYDROGENS by adding H +. DO NOT ADD ELECTRONS (This has already been factored in by the previous step) Ex. Balance P 4 H 2 PO 2 - + PH 3 in acidic solution Try: Balance Zn + As 2 O 3 AsH 3 + Zn 2+ in basic solution.

Try: Balance S 2- + ClO 3 - Cl - + S (basic) -17- Try: Balance CN - + IO 3 - I - + CNO - (acidic) Try: Balance As 4 + NaOCl + H 2 O AsO 4 3- + NaCl VI. REDOX TITRATIONS Acid-base titrations are very useful as they allow an accurate determination of an unknown concentration of an acid or a base. In a similar manner, there are many occasions where you may need to know the concentration of a substance that is capable of undergoing an oxidation or reduction reaction. A. OXIDIZING AGENTS One of the most useful oxidizing agents that you will encounter is acidic KMnO 4. The half reaction is: MnO 4 - + 8H + + 5e- Mn 2+ + 4H 2 O

-18- It has such a strong tendency to reduce (note its position on the table high on the top left) that it is able to oxidize a large number of substances (the K + in KMnO 4 is left out as it is a spectator ion). Ex. To find the [Fe 2+ ] in a unknown solution, react it with acidic MnO 4 - as follows: Recall that acid base titrations use an indicator to help see the equivalence point of the titration. The above redox titration also requires some way to identify the equivalence point. Another reason the KMnO 4 is so commonly used in redox titrations is that: MnO 4 - + 8H + + 5e- Mn 2+ + 4H 2 O Purple colourless the species being reduced also acts as The indicator. At the equilvalence pont The colour disappears. Ex. A 100.0 ml sample containing FeCl 2 is titrated with 0.100 M KMnO 4 solution. If 29.15 ml of KMnO 4 was required to reach the endpoint, what was the [Fe 2+ ]? 1. Balance redox 2. Calculate moles MnO 4-3. Use mol ratio to calculate Fe 2+ 4. Determine [ ] B. REDUCING AGENTS Two commonly used reducing agents are NaI and KI. A large number of substances can oxidize I - to I 2 (as it is relatively low about halfway down the table) according to the following half-reaction: 2I - I 2 + 2e- Titrations involving I - generally involve two consecutive steps: 1. First the I - is oxidized to I 2 by the substance being reduced. 2. Secondly, the I 2 produced in step 1 is reduced back to I - by a second reducing agent such as the thiosulphate ion (S 2 O 3 2- )

An example of a reaction involving I - is the reduction of laundry bleach, NaOCl. The reaction between I - and OCl - proceeds as follows: -19- No attempt is made to add exactly enough I - to react with the OCl -. Rather: A deliberate excess of I - is added on order to ensure that all the OCl - has reacted. ([OCl - ] is what is being looked for). Any excess I- will not affect the results. The important part is that every molecule of OCl - has reacted to form a molecule of I 2 (or as much as possible) The above reaction between OCl - and I - is the initial reaction because the actual redox titration involves a second reaction between the I 2 produced, and another ion present in the titrating substance, the reducing agent sodium thiosulphate, Na 2 S 2 O 3. (which ionizes into sodium and thiosulphate) When the addition of S 2 O 3 2- has reacted most of the I 2 present, the brown colour of the I 2 almost disappears (a diluted colour appearing pale yellow remains). Some starch solution is then added to the titration, which produces a dark blue colour (this is caused by the reaction between starch and the remaining I 2 in solution). After adding the starch (which acts as a more noticeable and therefore more precise indicator), the last of the S 2 O 3 2- is added, causing the blue colour of the starch-i 2 mixture to fade so that the last of the colour just disappears at the equivalence point between the I 2 and the S 2 O 3 2-. This whole process is done since there is no suitable indicator for the reaction between OCl - and I -, so a substance is reacted that can then undergo another reaction that has a suitable indicator. Ex. A 25.00 ml sample of bleach is reacted with excess KI according to the following equation: 2 H + + OCl - + 2 I - Cl - + H 2 O + I 2 The I 2 produced requires exactly 46.84 ml of 0.7500 M Na 2 SO 3 to bring the titration to the endpoint using starch solution as an indicator, according to the following equation: 2-2- 2 S 2 O 3 + I 2 S 4 O 6 + 2 I - What is the [OCl - ] in the bleach?

-20- VII. ELECTROCHEMICAL CELLS Recall the half-reactions need to be somehow connected in order for both reactions to occur (a donating and accepting of electrons needs to occur). ELECTROCHEMICAL CELL A system of half reactions joined together in such a way as to PRODUCE ELECTRICAL ENERGY Consider the reaction: Cu 2+ + Zn Cu + Zn 2+ A spontaneous reaction will occur when zinc metal is placed into a solution of CuSO 4. However, The electrons would be transferred directly from the Zn to the copper, being used up as the reaction proceeds. The exact same reaction can be used to produce electricity if: the half reactions are kept separate in such a way as to allow the reactions to still occur. ELECTRODE A conductor at which a half reaction occurs. (A general term) ANODE The electrode at which OXIDATION occurs. It is the electrode receiving the electrons from a substance being oxidized. All anions in the system travel towards the anode.

CATHODE The electrode at which REDUCTION occurs. It is the electrode supplying electrons to a substance being reduced. All cations travel towards the cathode. -21- Memory aid: Oxidation occurs at the Anode Reduction occurs at the Cathode (vowels) (constonants) EXAMPLE ELECTROCHEMICAL CELL DIAGRAM 1.The possible half-reactions are: Ag Ag + + e- Cu Cu 2+ + 2e- Before connecting, each cell has a small tendency to react (lose electrons and prduce positive ions), such that any electrons given off remain on the metal. (Not to a great extent) Experimentally, Cu has a greater tendency to oxidize. 2. After the half-cells are connected: Cu has a greater tendency to oxidize, causing an excess of electrons to accumulate on the copper electrode. This excess of electrons causes the electrons to flow from the Cu to the Ag electrode, but by wires. As a result, the Cu loses electrons (oxidizes) and Cu becomes the ANODE. THE ANODE IS WHERE OXIDATION OCCURS 3. As electrons are supplied to the Ag electrode: (via the wire) The equilibrium Ag Ag + + e- is upset, and according to Le Chataliers principle, the Ag + will be reduced to Ag. A reduction occurs at the Ag electrode, and it becomes the cathode. THE CATHODE IS WHERE REDUCTION OCCURS

4. Overall, the electron flow is: -22- from the Cu electrode (oxidation is producing electrons) to the Ag electrode (reduction is using the electrons) 5. The salt bridge: Prevents free mixing of the solutions. Water and certain ions are able to pass through the salt bridge, but are blocked somewhat. Ag + ions are reduced to Ag(s) in one half-cell while being unable to contact the Cu directly. (If it did, there would be direct transfer of electrons so they would not flow through the wire and no electrical energy could be harnessed,) By separating the half cells, electrons are forced to flow through the wire. After time, ions begin to move through the salt bridge to balance out the charge difference. A device such as a voltmeter is attached along the wire to measure the flow of electrons. 6. As Cu 2+ ions are formed: ELECTRONS AWAYS FLOW FROM ANODE CATHODE (A C) They accumulate around the anode. This excess positive charge is depleted by two simultaneous migrations. 1. Cu 2+ move away from the anode (random movement (greater probability it will move away from the high [Cu 2+ ] around the anode than approach it) 2. Negative ions (such as the SO 4 2- ions in sol n and the negative ions in the salt bridge move towards the anode. Due to the positive ions produced at the anode, they will attract the negative ions towards it. 7. As the [Ag + ] is depleted around the cathode: (as it is being reduced to Ag(s)) The net amount of positive charge is decreased near the cathode. The deficiency of positive charge is depleted by two simultaneous migrations: 1. Ag + ions move towards the cathode (random movement probability) as they positive ions are depleted near it. 2. Negative ions (such as the NO 3 - in solution) move away from the cathode. The greater amount of positive charge at the anode end of the cell will attract the negative ions and they will begin to flow to it (into the salt bridge) NOTE 1. NO electrons flow in sol n, only ions. The electrons flow through the wire. 2. The number of electrons involved in the oxidation reaction must equal the number of electrons involved in the reduction reaction.

-23- Ex. Assume two half-cells consisting of Pb(s) in a Pb(NO 3 ) 2 solution and Zn(s) in a ZnCl 2 solution are connected to make an electrochemical cell. Draw and label the parts of the cell, write the equations for the individual half reaction and overall reaction, and indicate the direction in which the ions and electrons move. 1. write possible ½ rxns 2. determine ox/red 3. write balanced rxn 4. draw cell and label

VIII. STANDARD REDUCTION POTENTIALS -24- VOLTAGE The tendency of electrons to flow in an electrochemical cell is called the VOLTAGE or ELECTRICAL POTENTIAL to do work. The voltage is the WORK DONE PER ELECTRON TRANSFERED. Since electrons cannot flow in an isolated half-cell, the voltage of an individual half-cell cannot be determined. However: The difference in electrical potential between two half-reaction can be measured A ZERO-POINT is arbitrarily defined on the voltage scale. Specifically the voltage for the HYDROGEN HALF-CELL: Defined as 2H + (aq) + 2e- H 2 (g) E o = 0.00 V E o = the STANDARD REDUCTION POTENTIAL in volts. The o in E o indicates it is at standard state. (If it is not, it will just state E =) An electrochemical cell is said to be at STANDARD STATE if: 1. It is at 25 o C 2. All gases are at 101.325 kpa (1 atm) 3. All elements are in their standard states (normal phase at 25 o C) 4. All solutions involved in the half-cell (both reactants and products) have a concentration of 1.0 M ALL of these must be true All voltages listed in the table of STANDARD REDUCTION POTENTIALS are determined at standard state and are compared to the standard reduction potential of the hydrogen half-cell. Ex. Cu 2+ + 2e- Cu E o = + 0.34 This half-cell has a voltage 0.34 more than the hydrogen half-cell. Ex. Zn 2+ + 2e- Zn E o = -0.76 This half-cell has a voltage 0.76 less than the hydrogen half-cell.

Since the voltage is a measure of work done (so work is either done or is being done), reversing a reduction reaction such as -25- Zn 2+ + 2e- Zn E o = -0.76 V produces: an oxidation reaction with a CHANGED SIGN for E o. Zn Zn 2+ + 2e- E o = +0.76 V IF A HALF REACTION IS REVERSED, THE SIGN OF ITS E o VALUE IS ALSO REVERSED. When two half reactions are combined, The voltage for the overall reaction is found to be the difference between the voltages of the individual half cells (one reduction one oxidation) Ex. Hg 2+ + 2e- Hg E o = +0.85 V Cu 2+ + 2e- Cu E o = +0.34 V Since two half cells are added to give the redox reaction, the voltages can also be added. E o (cell) = E o (reduction) + E o (oxidation) ONCE you have adjusted the signs. Ex. Calculate the potential of the cell Ni 2+ + Fe Ni + Fe 2+

-26- If E o cell is positive for a redox reaction, the reaction is expected to be spontaneous. If E o cell is negative, then the reaction is non-spontaneous. Ex. Calculate the potential of the cell: Mn + Mg 2+ Mn 2+ + Mg Mn 2+ + 2e- Mn Mg 2+ + 2e- Mg E o = -1.19 V E o = -2.37 V Mg 2+ + 2e- Mg E o = -2.37 V Mn Mn 2+ + 2e- E o = +1.19 V E o (cell) = -1.18 V NOT SPONTANEOUS Ex. Calculate the potential of the cell: 3 Ag + + Al 3Ag + Al 3+ Try: Calculate the potential for the cell and state if it would be expected to be spontaneous: 2 H 2 O 2 2 H 2 O + O 2 Although E o can be used to predict if a reaction is spontaneous, it has no correlation to the rate of the reaction. Recall that the activation energy of a reaction determines the rate at which it will proceed.

COMMENT ON WATER -27- Sometimes the reduction of neutral water is shown as: 2H + (10-7 M) + 2e- H 2 (g) E o = -0.41 V Other times it is shown as: 2H 2 O + 2e- H 2 (g) + 2OH - (10-7 M) E o = -0.41 V If a reaction occurs in a neutral solution, the reduction of neutral water (at 0.41 V) may be a possible reaction and must be considered along with any other reductions possible. If a reaction occurs in an acidic solution, the reduction of H + (at 0.00 V) may be a possible reaction and must also be considered along with any other reductions possible. 2H + + 2e- H 2 (g) E o = 0.00 V (you won t be asked to deal with basic solutions) SURFACE AREA OF ELECTRODES The surface area of electrodes has NO EFFECT on the cell potential. Increasing the surface area will increase the rate of the reaction (number of electrons per second amps) but doesn t change the voltage (work done by each). Increasing the electrode surface area will also increase the length of time that the cell can operate. Also, since the electrodes are solids, changing the size of the electrodes will not affect the cell (solids have constant concentration) HALF-REACTIONS NOT AT STANDARD STATE If a half-cell is not at standard conditions, there will be a change in the potential. Ex. Cu 2+ + 2e- Cu E o = +0.34 V

-28- According to Le Chatalier s principle, if the [Cu 2+ ] so it is > 1.0 M, the equilibrium will shift To the right, so the reduction potential increases. Cu 2+ + 2e- Cu E = > 0.34 V And if the [Cu 2+ ] is decreased so it is < 1.0 M, the equilibrium will shift: To the left, so the reduction potential decreases Cu 2+ + 2e- Cu E = < 0.34 V Notice that the o is dropped from the above equations as they are NOT at standard state. CELLS THAT REACH EQUILIBRIUM Operating chemical cells ARE NOT AT EQUILIBRIUM. The reaction arrow is in one direction. Ex. 2Ag + + Cu 2Ag + Cu 2+ Inutually, great tendency to form products. As rxn Proceeds, [ ]s change (rxt used, prod formed) THE REDUCTION REACTION (2Ag + + 2e- 2Ag) As the [Ag + ] decreases as the rxn proceeds, the reduction potential decreases and the half reaction lowers, relative to the chart. Due to the decrease in concentration, the tendency to form products decreases as the cell operates. THE OXIDATION REACTION (Cu 2+ + 2e- Cu) As the [Cu 2+ ] increases, the tendency for the reaction to undergo reduction increases. Therefore the tendency to be oxidized is increasingly opposed by a growing tendency to be reduced as the reaction proceeds. Overall, the following occurs as cell go towards equilibrium: INITALLY AS RXN PROCEEDS EVENTUALLY (TIME PASSES)

-29- The reduction potential of the reduction reaction decreases and the reduction potential of the oxidation reaction increases as the reaction proceeds until both half-reactions have the same reduction potential. Eventually the difference reaches zero at this point it is at equilibrium. IX. SELECTING PREFERRED REACTIONS When a cell contains a mixture of substances, several different half-reactions (and therefore overall reactions) may appear to be possible. Consider the following cell: The possible half-reactions are: Ag + + e- Ag(s) Cu 2+ + e- Cu(s) Zn 2+ + e- Zn(s) E o = +0.80 V E o = +0.34 V E o = -0.76 V WHEN SEVERAL DIFFERENT REDUCTION HALF-REACTIONS CAN OCCUR The half-reaction having the highest reduction potential will occur preferentially. (THE STRONGEST RA the highest one WILL REACT FIRST) Ag + will reduce before the other two and form Ag(s) WHEN SEVERAL DIFFERENT OXIDATION HALF-REACTIONS CAN OCCUR The half-reaction haiving the lowest reduction potential will oxidize. (THE STRNGEST OA the lowest one WILL REACT FIRST) Zn will oxidize before the others and form Zn 2+ TO DETERMINE THE PREFERRED HALF-REACTIONS: 1. IDENTIFY ALL THE SPECIE PRESENT AND LOCATE ON CHART (look at both sides) 2. IDENTIFY THE STRONGEST RA (highest on left side) IT WILL REACT FIRST 3. IDENTIFY THE STRONGEST OA (lowst on left side) IT WILL REACT FIRST

-30- Ex. A strip of iron metal is placed in a mixture of Br 2 (aq) and I 2 (aq). What is the preferred reaction that will occur? SPECTATOR IONS Any ion capable of being reduced will be a spectator ion if: There is another ion in the same solution that has a greater tendency to be reduced. Any ion capable of being oxidized will be a spectator ion if: There is another ion in the same solution that has a greater tendency to be oxidized. Some ions are more commonly used in making electrochemical cells because they have such a low tendency to oxidize or reduce that they rarely come into play in a reaction. Common Spectator Ions: Na + K + Ca 2+ Mg 2+ SO 4 2- (in neutral sol n) Cl - X.CORROSION OF METALS CORROSION Undesired oxidation of metals other than iron. (Getting eaten away ) (Undesired oxidation of iron = rusting) When a drop of water rests on an iron surface, a spontaneous reaction occurs: At the oxygen-poor region in the center of the drop, the iron oxidizes. Fe(s) Fe 2+ + 2e-

-31- Once the Fe 2+ is formed, it tends to migrate away from the anode (random movement from higher to lower concentrations). As they move away, more Fe(s) is exposed underneath the drop. At the same time, the reaction: ½ O 2 + H 2 O + 2e- 2OH - Is occurring at the oxygen-rich outer surface of the drop. When the Fe 2+ reaches this region, it encounters the OH - and precipitates as Fe(OH) 2 (s). The Fe(OH) 2 is eventually oxidized to a complex mixture of Fe 2 O 3 and H 2 O by the O 2 in the air. Rust is Fe 2 O 3. XH 2 O where X can change. Rust can have numerous different colours (red, brown, yellow, black) since differing numbers of water molecules attached to the iron(iii) oxide will change the colour of the compound. A metal can corrode if it touches a different type of metal in the presence of an electrolyte solution exposed to oxygen. For example, if iron touches copper wire and the spot where they touch gets wet, then: Fe Fe 2+ + 2e- (Fe has a greater tendency to oxidize than Cu) The copper conducts the electrons away from the Fe and makes them available to the oxygen/water touching the wire. ½ O 2 + H 2 O + 2e- 2OH - PREVENTING CORROSION There are several ways to stop or at least slow down corrosion, all of which fall into two main categories. 1. ISOLATING THE METAL FROM ITS ENVIRONMENT a. Apply a protective coating (paint or plastic) to the surface. If oxygen and water can t get to the metal, it won t corrode. b. Apply a metal which is corrosion resistant to the surface of the original metal. Ex. Tin cans steel is plated with tin (steel strong, tin oxidizes and forms a strong protective coating of tin oxide which adheres strongly to the steel underneath, preventing further corrosion) 2. ELECTROCHEMICAL METHODS a. Cathodic protection The process of protecting a substance from unwanted oxidation by connecting it to a substance having a higher tendency to oxidize.

Ex. Both Mg and Zn are stronger reducing agents than iron, so if pieces of magnesium or zinc are attached to the surface of iron, the Mg or Zn will: -32- be preferentially reduced and act as the anode. That will force the Fe to act as a cathode relative to the Mg (Fe remains in it s reduced form) Ex. Ex. Ex. Strips of zinc are often bolted to the iron-hull of ships below the water line. The zinc is sacrificed (and eventually needs to be replaced) to keep the iron of the ship from oxidizing and corroding. Some ships even pass a low voltage electric current into the hull from an electric generator. This forces electrons into the metal and prevents it from being oxidized. Galvanized iron simply has a zinc coating. The zinc reacts preferentially with air and water, forming a zinc oxide layer which protects the iron as it adheres strongly and prevents the exposure of the underlying metal to air and water. Some buried gas and oil tanks made of steel have a thick braided wire connected to them. The wire comes to the surface and is attached to a post in the ground that is made of an easily oxidized metal such as magnesium or zinc. Because the post will oxidize first, the buried tank is protected. b. Change the conditions of the surroundings When iron is placed in contact with water containing oxygen, the following halfreaction will oxidize the iron: ½O 2 + 2H + (10-7 M) + 2e- H 2 O E o = +0.82 V If oxygen is removed from the system, the tendency for it to reduce is drastically reduced. (Hydrogen will be reduced first as it is higher on the table). 2H + (10-7 M) + 2e- : H 2 (g) E o = -0.41 V Another method is to lower the [H + ] ions by adding OH - ions. A piece of iron will not rust (or a very little amount until all the oxygen has reacted) in a basic solution. XI. ELECTROLYSIS ELECTROLYSIS The process of supplying electrical energy to a molten ionic compound or a solution containing ions so as to produce a chemical change. Supplying E to a non-spontaneous redox reaction, allowing them to occur. Called an electrolytic cell.

-33- ELECTROLYSIS OF A MOLTEN BINARY SALT BINARY SALT A salt made up of only two different elements. NaCl KBr MgI 2 AlF 3 When such a salt is melted, the ions are free to move in the liquid form (molten NaCl is NaCl(l), as only Na + and Cl - ions are present. Do not confuse it with NaCl(aq) which is a solution containing Na +, Cl -, and H 2 O.) Also note that there is no need for a salt bridge to keep the reactant separated as no spontaneous reaction will occur between the reactants. The only reactants present are Na + and Cl - OA RA The anode reaction is: 2Cl - Cl 2 + 2e- E o = -1.36 V The cathode reaction is: Na + + e- Na E o = -2.71 V The overall reaction is: 2Na + +2Cl - Cl 2 + 2Na E o = -4.07 V Recall: non-spontaneous reactions have a negative voltage (and the OA is below the RA on the chart) In order for the above cell to operate: AT LEAST + 4.07V must be added Since the half cells are not at standard state, the reduction potentials will be different than those listed on the table.

ELECTROLYSIS OF AN AQUEOUS SOLUTION -34- Consider the electrolysis of aqueous sodium iodide, NaI(aq). This involves another consideration now water is also present in the system. Inert electrodes are used and the cell is set up as follows: During the electrolysis of aqueous solutions you must always consider the possibility that H 2 O may either oxidize and/or reduce. There are two possible oxidations: There are two possible reductions: So, in order to determine which reaction will occur, think back to the definition of electrolysis electrical energy is applied to produce a chemical change. It makes sense that the reaction that: Requires the least voltage input will be preferred (will occur first). Just as before, the strongest OA and the strongest RA will react first. Looking at the above example. The preferred reaction (requiring the least voltage input) involves the higher of the two possible reductions and the lower of the two possible oxidations. The same situation applies to electrolysis as did for spontaneous reactions: The half-reactions having the greatest tendency to reduce and greatest tendency to oxidize are preferred.

The half-reactions and overall reaction for the electrolysis of NaI(aq) is: -35-2H 2 O + 2e- H 2 (g) + 2OH - (10-7 M) E o = -0.41 V (cathode) 2I - I 2 + 2e- E o = -0.54 V (anode) 2H 2 O + 2I - H 2 (g) + 2OH - (10-7 M) E o cell = -0.95 V Min 0.95 V must be input Note The concentrations of the materials in cells in not relevant as long as there is sufficient material in the cell you can assume that the reactions proceed as predicted. In NEUTRAL AQUEOUS SOLUTIONS there are two equations involving water that must be considered: Oxidation ½ O 2 (g) + 2H + (10-7 M) + 2e- H 2 O E o = 0.82 V Reduction 2H 2 O + 2e- H 2 (g) + 2OH - (10-7 M) E o = -0.41 V In ACIDIC SOLUTIONS there are also two equations involving water that must be considered: Oxidation ½ O 2 (g) + 2H + + 2e- H 2 O E o = 1.23 V Reduction 2H + + 2e- H 2 (g) E o = 0.00 V Basic solutions will not be used with cells in Chemistry 12, as OH - can often precipitate metals out of solution. IN REALITY It is most often found that a higher potential than calculated must be applied to cause electrolysis. This is due to: Factors such as internal resistance and others. The difference between actual potentials required for electrolysis and the calculated potentials are termed overpotentials. As a result of the overpotential effect, when dilute neutral solutions ( <1.0 M) containing Cl - or Br - are electrolyzed:

Oxidation: Cl 2 + 2e- 2Cl- Br 2 + 2e- 2Br- ½ O 2 + 2H + (10-7 M) + 2e- H 2 O E o = -1.36 V E o = -1.09 V E o = -0.82 V -36- You would expect that O 2 would be produced. In reality it is found that Cl 2 or Br 2 are actually produced. This is indicated on the table the lines that state overpotential effect When dilute solution of certain metals are electrolyzed: Reduction: 2H 2 O + 2e- H 2 (g) + 2OH - (10-7 M) Fe 2+ + 2e- Fe Cr 3+ + 3e- Cr Zn 2+ + 2e- Zn E o = -0.41 V E o = -0.45 V E o = -0.74 V E o = -0.76 V You would expect that H 2 is produced, but in practice the metals are produced. Electrolysis of aqueous solutions containing Cl - or Br - will produce Cl 2 or Br 2 at the anode Electrolysis of aqueous solutions containing Fe 2+, Cr 3+, or Zn 2+ will produce Fe, Cr, or Zn at the cathode Ex. What products are formed at the anode and cathode and what is the overall reaction when a solution containing NiSO 4 (aq) is electrolyzed using inert electrodes? Determine the minimum voltage required. 1. detemine species present 2. determine strongest OA/RA 3. determine overall reaction 4. calculate answer

Ex. -37- What is the overall reaction which occurs when a 1.0 M solution of HCl(aq) is electrolyzed using carbon electrodes? XII. ELECTROPLATING ELECTROPLATING Electroplating is an electrolytic process in which a metal is reduced or plated out at a cathode The Cathode is: the site where plating occurs, so it is the material that will receive the metal plating. The Electroplating Solution is: a solution containing ions of the metal which is to be plated onto the cathode. The Anode is: made of either an inert material or made of the same metal which is to be plated onto the cathode. You will only be asked about electroplating in NEUTRAL solutions.

Ex. Design a cell to electroplate a copper medallion with nickel metal. Include in the design: a. the ions present in the solution b. the direction of ion flow c. the substance used for the anode and cathode d. the direction of electron flow when the cell is connected to a DC power source -38- cathode the medallion is to have Ni plated onto it, so it must be the cathode. anode can make it the same material to be plated, so often it is Ni to provide a supply of Ni 2+. ions must have Ni 2+ in sol n to provide nickel to plate onto the cathode. NO 3 - is used as a spectator. Ni ] flow towards the ve cathode, plating onto it. CATIONS FLOW TOWARDS CATHODE ANIONS TOWARDS ANODE Electrons flow A C ELECTROREFINING The process of purifying a metal by electrolysis At the anode the small amounts of Zn or Pb present is preferentially oxidized as it is exposed at the surface. When any exposed Pb/Zn atoms have oxidized and gone into solution as ions, only the Cu atoms are available to be oxidized. Any Au, Ag, or Pt atoms present cannot be oxidized because the anode is mostly copper which is oxidized in preference to Au, Ag, or Pt which simply drop off and accumulate on the bottom of the cell. This anode sludge can be purified to obtain the valuable metals.

XIII. APPLIED ELECTROCHEMISTRY -39- A. THE BREATHALYSER When alcohol is consumed, it is absorbed into the blood stream from the stomach. Some of it passes through the cell walls of blood capillaries into the alveoli (air sacs that make up the lungs), in a similar manner that CO 2 (g) and O 2 (g) enter in and out of the lungs. The process of alcohol entering the bloodstream and entering the lungs is at equilibrium so that the greater the concentration of alcohol in the blood, the greater the concentration of alcohol in the lungs. When the lungs exhale, the ethanol in the lungs is expelled. Ethanol undergoes oxidation by an acidic solution of dichromate ions as follows: 3 C 2 H 5 OH + 2 K 2 Cr 2 O 7 + 8 H 2 SO 4 3 CH 3 COOH + 2 Cr 2 (SO 4 ) 3 + 2 K 2 SO 4 + 11 H 2 O (Cr 2 O 7 2- is orange/yellow) (Cr 3+ is dark green) The expelled air is blown into a breathalyzer which can measure and record the amount of green colourization. The more green the more alcohol in the breath. The machine is calibrated to be able to accurately calculate the blood alcohol content. B. BATTERIES 1. The Lead-Acid Storage Battery A car battery is of this type, consists of alternating pairs of plates of Pb(s) and PbO 2 (s) immersed in dilute sulphuric acid. The anode reaction is: Pb(s) + HSO 4 - (aq) PbSO 4 (s) + H + (aq) + 2e- (or Pb Pb2+ + 2e-) The cathode reaction is: PbO 2 (s) + HSO 4 - (aq) +3H + +2e- PbSO 4 (s) + 2H 2 O (or Pb 4+ + 2e- Pb 2+ ) The overall reaction (Pb + Pb 4+ Pb 2+ + Pb 2+ ) occurs when the battery is discharging spontaneously reacting to produce electrical energy. This forms an insoluble PbSO 4 (s) layer on the plates of the battery. When an external charge is applied, the reaction is driven backwards. (Over time the PbSO 4 tends to flake off the plates so less Pb and PbO2 can be formed, so eventually the battery needs to be replaced.)

2. The alkaline battery -40- Gets its name from the alkaline (basic) electrolyte (KOH) that is used. The cathode reaction: 2 MnO 2 (s) + H 2 O(l) +2e- Mn 2 O 3 (s) + 2 OH - (aq) The anode reaction: Zn(s) + 2OH - (aq) ZnO(s) + H 2 O(l) +2e- Cheap to make, but cannot be recharged (or reversed), has a relatively short shelf life (reaction continually occurs slowly whether the battery is being used or not).