5.9 Standard Reduction Potentials and Voltages

Size: px

Start display at page:

Download "5.9 Standard Reduction Potentials and Voltages"

Joella Jackson
5 years ago
Views:

1 5.9 Standard Reduction Potentials and Voltages Voltage (or Electrical Potential):.( p. 324) Can also be defined as the potential energy per coulomb. (Where 1 C = the charge carried by 6.25x10 18 e - ) 1 Volt = 1 Joule/Coulomb. Reduction potential of half-cells - The tendency of a half-cell to be reduced. (take e - s) Voltage only depends on the difference in potentials Question: Are you tall? Yes/No Height Height Compared to Kobe Yao Ming cm Kevin Garnett Kobe Bryant Steve Nash You?? Note: Tall is a relative term. Tall in the NBA tall in your classroom On the Standard Reduction Table all half reactions are compared to the reduction of 2 H + (aq) + 2 e - H 2(g) E o = v this half-cell was arbitrarily chosen to compare other half-cells with. It was assigned a reduction potential of v Note: Relative potentials of half-cells can only be determined by connecting with other half-cells and reading the voltage E o (Standard Conditions) Recall: the rate of a chemical reaction can be affected by many factors (see p. 6 of Hebden) certain conditions were used to determine E o temperature = gas pressure = elements are concentration of all solutions involved = o e.g. 2 H + (aq) + 2 e - H 2(g) E o = v o at standard conditions [H+] = [H 2 ]= 1

2 Determining the voltage of a cell An electrochemical cell requires 2 Voltage of the cell = the sum of the potential difference of the 2 half reactions If you reverse a half reaction you must change the sign of the potential difference o Zn e - Zn ; E o = V o Zn Zn e - ; E o = Steps Required 1) Find the two metals on the reduction table. Higher one is the cathode. (HIC) 2) Write the cathode half-rx as is on the table (cathode - reduction) Include the reduction potential (E o ) beside it. 3) Reverse the anode reaction (AIR) (anode oxidation) Reverse the sign on the E o (If (+) (-) If (-) (+)) 4) Multiply half-rx s by factors that will make e - s cancel. DON T multiply the E o s by these factors. 5) Add up half-rx s to get overall redox reaction. 6) Add up E o s (as you have written them) to get the initial voltage of the cell. E.g.1) Ag + /Ag with Cu 2+ /Cu cathode half-rx: ;E o = anode reaction ;E o = Overall rxn: ; E o CELL= E.g. 2) What is the potential of the following cell? Ni 2+ + Fe Ni + Fe 2+ cathode Ni e - Ni ; E o = anode Fe Fe e - ; E o = Ni 2+ + Fe Ni + Fe 2+ ; E o CELL = cathode Ni e - Ni anode Fe Fe e - Or you could use the following formula E o CELL = E o RED - E o OX = (-0.26) - (-0.45) = 2

3 E.g. 3 Cathode half-rx Ag + + e - Ag E o = Anode half-rx H 2 2H + + 2e - E o = add half rxns 2Ag + + H 2 2H + + 2Ag E o CELL = Note: Ag + + e - Ag = V 2 Ag + +2 e - 2 Ag = V Why? Voltage = work Electron Voltage = 2 x work 2 electrons LEOA GERC E.g. 4 Calculate the voltage of Ni + Fe 2+ Ni 2+ + Fe anode Ni Ni e - cathode Fe e - Fe E o CELL = E o RED - E o OX = The sign means this reaction is non- spontaneous a reaction with a sign would be spontaneous Surface Area Question: How does surface area of the electrode affect the voltage of a cell (p.223)? What factor is affected by an increase or decrease in the surface area of the electrode? Standard vs Non-standard Voltage is calculated under If concentration increases from standard conditions then the reduction potential If concentration decreases from standard conditions then the reduction potential Note: you do not have to know why the above phenomena occur but you do have to know the result. 3

4 Reaching Equilibrium (p.224) Ni e - Ni Fe Fe e - ; E o = V ; E o = V initially concentrations = and voltage = as the cell operates 2 things occur 1. [Ni 2+ ] o Therefore tendency to form 2. [Fe 2+ ] at the anode o Therefore the tendency to be is increasingly opposed by tendency to be Eventually the cell will reach At this time the reduction potential of the 2 half cells will be And E o CELL = 5.10 Selecting Preferred Reactions Read p Ex # Using the Standard Reduction Table you can predict which reaction will occur and therefore the voltage that is produced. Draw another wire connected to an Ag electrode in the beaker on the right Cu Zn2+ NO 3 - Cu 2+ Ag + NO 3-4

5 The half reactions are Ag + +e - Ag (s) ; E o = Cu e - Cu (s) ; E o = Zn e - Zn (s) ; E o = Zn has to act as the Strongest agent But what will happen in the other beaker? Hint: which is the stronger oxidizing agent? Therefore will be the cathode What is the voltage from the cell above? (show half reactions and the redox reaction that will occur) Effects of Water on Electrochemical Cells the reduction of water can be written 2 ways (p.222) If water is present we need to consider this reaction as a preferential reaction Also if conditions are acidic we need to consider the reduction of H + as a preferential reaction 2 H e - H 2(g) ; E o = 0.00 V Spectator Ions Ions that are capable of being reduced will be spectator ions if another ion in the solution has a Ions that are capable of being oxidized will be spectator ions if another ion in the solution has a Generally these are spectator ions Na +, K +, Ca 2+, Mg 2+, SO 4 2- Read p. 222, 223, Ex #

6 5.9 Standard Reduction Potentials and Voltages Voltage (or Electrical Potential): the tendency of electrons to flow in an electrochemical cell: the work done per electron transferred.( p. 324) Can also be defined as the potential energy per coulomb. (Where 1 C = the charge carried by 6.25x10 18 e - ) 1 Volt = 1 Joule/Coulomb. Reduction potential of half-cells - The tendency of a half-cell to be reduced. (take e - s) Voltage only depends on the difference in potentials Question: Are you tall? Yes/No Height Height Compared to Kobe Yao Ming cm Kevin Garnett Kobe Bryant Steve Nash You?? Note: Tall is a relative term. Tall in the NBA tall in your classroom On the Standard Reduction Table all half reactions are compared to the reduction of H + 2 H + (aq) + 2 e - H 2(g) E o = v this half-cell was arbitrarily chosen to compare other half-cells with. It was assigned a reduction potential of v Note: Relative potentials of half-cells can only be determined by connecting with other half-cells and reading the voltage E o (Standard Conditions) Recall: the rate of a chemical reaction can be affected by many factors (see p. 6 of Hebden) certain conditions were used to determine E o temperature = 25 o C gas pressure = 1 atm (101.3 kpa) elements are in the phase that they exist at 25 o C concentration of all solutions involved = 1 M o e.g. 2 H + (aq) + 2 e - H 2(g) E o = v o at standard conditions [H+] = 1 M [H 2 ]= 1 M 6

7 Determining the voltage of a cell An electrochemical cell requires 2 half reactions Voltage of the cell = the sum of the potential difference of the 2 half reactions If you reverse a half reaction you must change the sign of the potential difference o Zn e - Zn ; E o = V o Zn Zn e - ; E o = V Steps Required 7) Find the two metals on the reduction table. Higher one is the cathode. (HIC) 8) Write the cathode half-rx as is on the table (cathode - reduction) Include the reduction potential (E o ) beside it. 9) Reverse the anode reaction (AIR) (anode oxidation) Reverse the sign on the E o (If (+) (-) If (-) (+)) 10) Multiply half-rx s by factors that will make e - s cancel. DON T multiply the E o s by these factors. 11) Add up half-rx s to get overall redox reaction. 12) Add up E o s (as you have written them) to get the initial voltage of the cell. E.g.1) Ag + /Ag with Cu 2+ /Cu cathode half-rx: Ag + + e - Ag anode reaction Cu Cu 2+ +2e - Overall rxn: ;E o = V ;E o = V 2 Ag + + Cu 2 Ag + Cu 2 ; E o CELL= 0.46 V E.g. 2) What is the potential of the following cell? Ni 2+ + Fe Ni + Fe 2+ cathode Ni e - Ni ; E o = V anode Fe Fe e - ; E o = V Ni 2+ + Fe Ni + Fe 2+ ; E o CELL = 0.19 V Or you could use the following formula cathode Ni e - Ni anode Fe Fe e - E o CELL = E o RED - E o OX = (-0.26) - (-0.45) = 0.19 V 7

8 E.g. 3 Cathode half-rx Ag + + e - Ag E o = Anode half-rx H 2 2H + + 2e - E o = 0.00 volts add half rxns 2Ag + + H 2 2H + + 2Ag E o CELL = 0.80 volts Note: Ag + + e - Ag = V 2 Ag + +2 e - 2 Ag = V Why? Voltage = work Electron Voltage = 2 x work 2 electrons LEOA GERC E.g. 4 Calculate the voltage of Ni + Fe 2+ Ni 2+ + Fe anode Ni Ni e - cathode Fe e - Fe E o CELL = E o RED - E o OX = (-0.45) - (- 0.26) = V The negative sign means this reaction is non- spontaneous a reaction with a positive sign would be spontaneous Surface Area Question: How does surface area of the electrode affect the voltage of a cell (p.223)? No affect What factor is affected by an increase or decrease in the surface area of the electrode? Current (Amperes) more electrons transferred Standard vs Non-standard Voltage is calculated under standard conditions If concentration increases from standard conditions then the reduction potential increases If concentration decreases from standard conditions then the reduction potential decreases Note: you do not have to know why the above phenomena occur but you do have to know the result. 8

Reaching Equilibrium (p.224) Ni 2+ + 2 e - Ni Fe Fe 2+ + 2 e - ; E o = - 0.26 V ; E o = +0.45 V initially concentrations = 1.0 and voltage = 0.19 V as the cell operates 2 things occur 1.

9 Reaching Equilibrium (p.224) Ni e - Ni Fe Fe e - ; E o = V ; E o = V initially concentrations = 1.0 and voltage = 0.19 V as the cell operates 2 things occur 1. [Ni 2+ ] decreases o Therefore tendency to form products decreases 2. [Fe 2+ ] increases at the anode o Therefore the tendency to be oxidized is increasingly opposed by tendency to be reduced Eventually the cell will reach equilibrium At this time the reduction potential of the 2 half cells will be equal And E o CELL = Selecting Preferred Reactions Read p Ex # Using the Standard Reduction Table you can predict which reaction will occur and therefore the voltage that is produced. Draw another wire connected to an Ag electrode in the beaker on the right Cu Zn2+ NO 3 - Cu 2+ Ag + NO 3-9

10 The half reactions are Ag + +e - Ag (s) ; E o = Cu e - Cu (s) ; E o = Zn e - Zn (s) ; E o = Zn has to act as the anode Strongest reducing agent But what will happen in the other beaker? Hint: which is the stronger oxidizing agent? Ag Therefore Ag will be the cathode What is the voltage from the cell above? (show half reactions and the redox reaction that will occur) Cathode Ag + +e - Ag (s) ; E o = V Anode Zn (s) Zn e - ; E o = V Ag + + Zn (s) Zn 2+ + Ag (s) E o CELL = 1.56 V Effects of Water on Electrochemical Cells the reduction of water can be written 2 ways (p.222) If water is present we need to consider this reaction as a preferential reaction Also if conditions are acidic we need to consider the reduction of H + as a preferential reaction 2 H e - H 2(g) ; E o = 0.00 V Spectator Ions Ions that are capable of being reduced will be spectator ions if another ion in the solution has a greater tendency to be reduced Ions that are capable of being oxidized will be spectator ions if another ion in the solution has a greater tendency to be oxidized Generally these are spectator ions Na +, K +, Ca 2+, Mg 2+, SO 4 2- Read p. 222, 223, Ex #

Sec 5.8 Electrochemical Cells

Sec 5.8 Electrochemical Cells Demonstration (Cu/Zn Cell) Cu Definitions 1 M Zn(NO 3 ) 2 1 M Cu(NO 3 ) 2 Electrochemical cell A device which converts chemical energy into electrical energy Electrode A conductor