Electrochemistry Crash Course Electrochemistry is essentially the study of reactions involving the transfer of electrons from one element to another or the study of systems that allow for the flow of voltage from one area to another, such as a battery in a circuit. These reactions can be classified as REDOX reactions (reduction and oxidation taking place). Reduction is the process in which an electron is gained by an element/compound and Oxidation is the process where an element/compound loses electrons. These reactions are grouped as REDOX reactions because for one element to gain an electron another needs to lose an electron. Consider the following system: Zn(s) + 2 HCl (aq) à ZnCl2 (aq) + H2(g) This reaction can be broken into two half- reactions, where we look more closely at the individual elements that are gaining/losing electrons in the reaction to go from an element to an ion or vice versa. We see that zinc goes from an elemental solid to form an ionic compound and in the process loses two electrons. We would say that the zinc is oxidized as it is giving up electrons in the reaction to the system. Zn à Zn 2+ + 2 e - At the same time the hydrogen ion in the hydrochloric acid is gaining an electron each to form elemental hydrogen gas. Since hydrogen is picking up electrons from the reaction system it is being reduced. 2 H + + 2 e - à H2 You will notice that in the above example the chlorine is ignored in the half- reactions because it remains as the chloride ion (Cl - ) throughout the reaction and is neither reduced nor oxidized. LEO says GER LEO: Loss of Electrons is Oxidation GER: Gain of Electrons is Reduction Two other terms that you will commonly see are oxidizing agent and reducing agent. The confusing part is that they seem to be the opposite of what we are calling them. An oxidizing agent is allowing oxidation to take place and is therefore the substance that is being reduced (Hydrogen from the example above). The reducing
agent is the substance allowing reduction to take place and is therefore oxidized (Zinc). For each of the following list which element is being oxidized and which is being reduced. Also explain which element is the oxidizing agent and the reducing agent. a) 4 Fe(s) + 3 O2(g) à 2 Fe2O3(s) b) 2 PbO(s) + C(s) à 2 Pb(s) + CO2(g) c) NiO(s) + H2(g) à Ni(s) + H2O(l) d) Sn(s) + Br2(l) à SnBr2(s) e) Fe2O3(s) + 3 CO(g) à 2 Fe(s) + 3 CO2(g) To write balanced half- reactions it is easiest to start with the full balanced equation and identify which elements are gaining/losing electrons. If you just know the elements that are gaining/losing, you may need to multiply one of the half- reactions to ensure that the number of electrons gained equals the number of electrons lost. Eg. Cu(s) à Cu 2+ (aq) + 2 e - Ag + (aq) + e - à Ag(s) In the above reaction copper is losing 2 electrons as it is oxidized to the copper (II) ion and silver is only gaining one electron as it is reduce from the silver ion to silver. To balance this system we would need two moles of silver for every mole of copper reacting. Cu(s) + 2Ag + (aq) + 2 e - à Cu 2+ (aq) + 2 e - + 2Ag(s) We can cancel out the common electrons on each side to get the net equation: Cu(s) + 2Ag + (aq) à Cu 2+ (aq) + 2Ag(s) Write a pair of balanced half- reactions for the following: a) Zn(s) + Cu 2+ (aq) à Zn 2+ (aq) + Cu(s) b) Mg(s) + 2 H + (aq) à Mg 2+ (aq) + H2(g) Oxidation States: An oxidation number is a positive or negative number corresponding to the oxidation state of an atom. In covalent compounds or polyatomic ions, the more electronegative element is treated as though it has complete control of the electron. An element that has lost an electron is given an oxidation number of +1 and an element that has gained an electron is given an oxidation number of - 1. In a neutral compound, the overall oxidation number must equal 0. In a polyatomic ion, the overall oxidation number must equal the charge of the ion. Below are some common rules to apply.
Eg. What is the oxidation number of the carbon in methane CH4? From the rules above, we know that hydrogen has an oxidation number of +1. We also know that this is a neutrally charged compound so the overall charge must equal 0. C + 4H = 0 C + 4(+1) = 0 C = - 4 So the oxidation number on carbon is - 4 Eg. 2 What is the oxidation number on manganese in permanganate ion, MnO4 -? Oxygen has a charge of - 2 and the overall ion has a charge of - 1, so: Mn + 4 O = - 1 Mn + 4(- 2) = - 1 Mn = +7 1) Determine the oxidation numbers of: a. S in SO2 b. S in SO4 2- c. I in MgI2 d. Cl in HClO4 e. Cr in Cr2O7 2- f. H in CaH2 2) Determine the oxidation number of nitrogen in: a. NaNO3 b. N2
c. NH4Cl The application/analysis of oxidation numbers can also help us determine whether an oxidation or reduction reaction has occurred. During a REDOX reaction electrons move from one element to another and change the oxidation number of the elements involved. If an element has an increase in oxidation number it was oxidized and if it has a decrease in oxidation number it was reduced. In the above example, sodium has an increase in oxidation number from 0 to +1 as it is oxidized to form the sodium ion and chlorine is reduced as it goes from an oxidation number of 0 to - 1. For each of the following reactions, assign oxidation numbers to each atom/ion and indicate whether the equation represents a redox reaction If it does, identify the oxidation and reduction. a) Cu(s) + 2 AgNO3(aq) à 2Ag(s) + Cu(NO3)2(aq) b) Pb(NO3)2(aq) + 2KI(aq) à PBI2(s) + 2KNO3(aq) c) Cl2(aq) + 2KI(aq) à 2KCl(aq) + I2(s) d) 2 H202(l) à 2 H2O(l) + O2(g) Balancing REDOX equations: Redox reactions have special rules for balancing to ensure that all of the electron transfers will be accounted for. To do this you will start by balancing each of the half- reactions and then multiplying them by factors that will allow the electrons to cancel out of the equation. The general steps are as follows: 1. Write the chemical formulas for the reactants and products 2. Balance all atoms, other than O and H 3. Balance O by adding H2O(l) 4. Balance H by adding H + 5. Balance the charge on each side by adding e - and cancel anything that is the same on both sides. For basic solutions there are a couple of extra steps 6. Add OH - (aq) on both sides equal to the number of H + present 7. Combine H + and OH - if they are on the same side to make H2O. Cancel out H2O that appears on both sides.
Eg. 1. Write the given reactants and products: 2. Look in the table of reduction potentials for the 2 half- reactions containing these substances, and arrange them so that Fe2+ and MnO4- appear on the left as reactants: 3. Balance the reactions, so that they each have the same number of free electron in the equation: Here we have multiplied the entire iron half- reaction by 5 to match the manganese half- reaction. 4. Find the global reaction by adding the 2 half- reactions: 1. Balance the following reactions in an acidic solution: a. Cu(s) + NO3 - (aq) à Cu 2+ (aq) + NO2(g) b. Mn 2+ (aq) + HBiO3(aq) à Bi 3+ (aq) + MnO4 - (aq) c. H2O2(aq) + Cr2O7 2- (aq) à Cr 3+ (aq) + O2(g) + H2O(l) 2. Balance the following reactions in basic solution: a. Cr(OH)3(s) + IO3 - (aq) à CrO4 2- (aq) + I - (aq) b. Ag2O(s) + CH2O(aq) à Ag(s) + CHO2 - (aq) c. S2O4 2- (aq) + O2(g) à SO4 2- (aq) Galvanic Cells: We are all familiar with batteries. These are cells that continuously convert stored chemical energy into electrical energy when a pathway is created to allow the flow of electrical energy. A Galvanic electric cells developed for scientific study. Within the galvanic cell are to half- cells, each composed of an electrode and an electrolyte.
In the above diagram you can see the two half cells (Copper/Copper ion and Hydrogen/Hydrogen Ion) separated by a salt bridge. The salt bridge allows ions (eletrolytes) to flow between solutions, essentially carrying electrons through the system. The galvanic cell spontaneously produces electricity as the electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). You will notice that the copper terminal in the cell above has a + sign, indicating that it is the strongest oxidizing agent, meaning it will accept the electrons given off by the hydrogen strip and the - sign on the anode signify that it is the reducing agent, giving electrons to the cell. The question is, how do we predict/know which substance in the compound with the be strongest oxidizing agent vs reducing agent? For this we need to go back to the activity series. You may recall that the activity series can be used to determine which metal solid will bump which aqueous metal ions from solution in a single displacement reaction. We find that, as all metals solids are reducing agents, that the higher the metal is situated on the activity series, the stronger reducing agent it is. Then, as all metal ions are oxidizing agents, the lower it appears on the activity series, the stronger oxidizing agent the ion would be.
The above cell is written in cell notation. In this notation we find that the oxidation component of the cell is written before the reduction component of the cell. As zinc is higher on the activity series, is will more readily donate electrons as the reducing agent and copper ion will more readily accept electrons as the oxidizing agent. Lets look at a system with silver and copper: Ag(s) Ag 2+ (aq) Cu 2+ (aq) Cu(s) As copper is higher on the activity series than silver, it will be the strongest reducing agent and the cathode (site of oxidation). The silver ion, being lower on the activity series, will be the strongest reducing agent, and accept electrons to form solid silver at the anode. 1. Indicate whether the following processes would occur at the cathode or the anode of a galvanic cell: a. Reduction half- reaction b. Oxidation half- reaction c. Reaction of the strongest reducing agent d. Reaction of the strongest oxidizing agent 2. For each of the following reactions in cell notation, identify the strongest oxidizing and reducing agents. Write chemical equations that would take place at the anode, the cathode and the net equation of the cell. a. Ag(s) Ag 2+ (aq) Zn 2+ (aq) Zn(s) b. Mg(s) Mg 2+ (aq) Cu 2+ (aq) Cu(s) Standard Cell Potentials: A standard cell is a galvanic cell where each of the half- cells contains half- reactions at SATP conditions and a concentration of 1.0mol/L for the aqueous entities. Through experimentation we have determined standard reduction potentials Er o for all metals. (See Appendix) This represents the ability of the half- reaction to attract electrons in a reduction half reaction. We can use these values to find the standard cell potential (ΔE o ) for the cell using the following equation:
ΔE o = Er o cathode - Er o anode Example: We have a system composed of aluminum and hydrogen electrodes shown by the following half- reactions: 2 H + (aq) + 2 e - à H2(g) Er o = 0.00 V Al 3+ (aq) + 3 e - à Al(a) Er o = - 1.66 V Since Aluminum is higher on the activity series it is the stronger reducing agent in the system and will be oxidized at the anode. Therefore, hydrogen will be reduced at the cathode. Cathode 3{2 H + (aq) + 2 e - à H2(g)} Er o = 0.00 V Anode 2{Al(a) à Al 3+ (aq) + 3 e - } Er o = 1.66 V Notice that we balance the electrons in the overall system by applying coefficients, but that these coefficients do not apply to the reduction potentials as they are not altered by these factors as electric potential represents energy per coulomb charge. So we get: ΔE o = Er o cathode - Er o anode = 0.00V (- 1.66V) = 1.66V 1. What is the Eo for the reaction Cl2 + 2I- 2Cl- + I2? 2. Find Eo for the reaction 4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2O 3. The net reaction of the mercury cell is: Zn + HgO + H2O Zn(OH)2 + Hg What substance is oxidized at the anode of this cell? 4. The overall reaction of the silver oxide battery, which is used in watches, may be written as Ag2O + Zn 2Ag + ZnO. The cathode of the battery is: a. a) Ag b. b) ZnO c. c) Ag2O d. d) Zn