Question

Predict the standard cell potential and calculate the standard reaction Gibbs free energy for galvanic cells having the following cell reactions:

a) 3 Zn(s) +2Bi^3+(aq)--> 3 Zn^2+(aq) + 2Bi(s)

b) 2 H2(g) + O2(g)--> 2 H2O(l) in acidic solution

c)2 H2(g) +O2(g)--> 2 H2O(l) in basic solution

d) 3 Au+(aq) --> 2 Au(s) + Au^3+ (aq)

Answer 1

a) 3 Zn(s) +2Bi^3+(aq)--> 3 Zn^2+(aq) + 2Bi(s)

E0cell = Ecathode - Eanode

        = 0.2 - (-0.763)

    = 0.963 v

DG0 = -nFE0cell = - 6*96500*0.963 = -557.577 kj

b) 2 H2(g) + O2(g)--> 2 H2O(l) in acidic solution

E0cell = Ecathode - Eanode

        = 1.229 -   0

        = 1.229 v

DG0 = -nFE0cell = - 4*96500*1.229 = -473.39 kj

c)2 H2(g) +O2(g)--> 2 H2O(l) in basic solution

E0cell = Ecathode - Eanode

        = 0.4 - (-0.8277)

        = 1.227 v

DG0 = -nFE0cell = - 4*96500*1.227 = -473.62 kj

d) 3 Au+(aq) --> 2 Au(s) + Au^3+ (aq)

E0cell = Ecathode - Eanode

        = 1.68 - (1.4)

        = 0.28v

DG0 = -nFE0cell = - 4*96500*0.28 = -108.08 kj