Question

These questions concern the oxidation-reduction reaction shown below

3Co2+(aq) + 2Cr(s) 3Co(s) + 2Cr3+(aq)

(a) Use standard reduction potentials from Appendix H in Harris to calculate the standard cell potential E° for the reaction.

V

(b) How many electrons are transferred in this cell's overall chemical reaction?

(c) From E°, calculate the value of G°, the Gibbs free energy change for the reaction under standard conditions.

kJ/mol

(d) From E°, calculate the value of the equilibrium constant Keq for the reaction.

Answer 1

3Co2+(aq) + 2Cr(s) ---------->3Co(s) + 2Cr3+(aq)

2Cr(s) ---------------> 2Cr^3+ (aq) + 6e^-                E0 = 0.74v

3Co^2+ (aq) + 6e^- -------> 3Co(s)                       E0   = -0.29v

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3Co2+(aq) + 2Cr(s) ---------->3Co(s) + 2Cr3+(aq)    E0 cell = 0.45v

b. 6 electrons are transferred in this cell's overall chemical reaction

n = 6

c. G⁰      = -nE0cell*F

                   = -6*0.45*96500

                   = -260550J/mole   = -260.55KJ/mole

d. G⁰      = -nE0cell*F

                   = -6*0.45*96500

                   = -260550J/mole   = -260.55KJ/mole

G⁰      = -RTlnK

-260550   = -8.314*298*2.303logK

-260550    = -5705.84logK

logK         = -260550/-5705.84

logK           = 45.6

K                = 10^45.6   = 3.98*10^45