Question

Consider a voltaic (galvanic) cell with the following metal electrodes. Identify which metal is the cathode and which is the anode, and calculate the cell potential. (Use the table of Standard Electrode Potentials.) (a) Cd(II) and Sc(III) Cathode: Cd(II) Sc(III) Anode: Cd(II) Sc(III) Ecell 42 (b) Pb(II) and In(III) Cathode: Pb(II) In(III) Anode: Pb(II) In(III) Ecell - (c) Ni(II) and Zr(IV) Cathode: Ni(II) Zr(IV) Anode: Ni(II) Zr(IV) Ecell- Supporting Materials

Answer 1

ANSWER

1. In Galavanic cell, Anode have Oxidation And Cathode have Reduction

a) Cd(2) and Sc(3)   

Cd(2) will be cathode and Sc(3) will be anode due to its reduction potential value,

According to Nernst Equation, at T=25°C

Ecell = E°cell - 0.0591/n × log[anodic conc. ]/[Catholic conc.]

Ecell = E°cell - 0.0591/6 ×log[Sc^3+]^2/[Cd^2-]^3

b) Pb(2) and In(3)

Here, Pb(2) will be anode and In(3) will be cathode due to reduction potential value,

Ecell = E°cell - 0.0591/6 ×log[Pb^2+]^3/[In^3-]^2

c) Ni(2) and Zr(4)

Ni(2) will be cathode and Zr(4) will be anode.

Ecll = E°cell - 0.0591/4 × log[Zr^4+]/[Ni^2-]^2

2. a) Standard reduction potential value, Mg^2+ = -2. 372Vand Au^+ = 1.498V

Mg(2) will be anode and Au(3) will be cathode

When reduction potential of both electrode taken to be account, E°cell = E°cathode - E°anode

E°cell = 1.498 -(-2.372) = 1.498+2.372=3.870V

E°cell = 3.870V

b) Over all Balance equation, Balance equation multiply with

Mg(s) === Mg^2(aq)+ + 2e-.......×3

Au^3(aq)+ 3e- === Au(s) ............ ×2

3Mg(s) + 2Au^3+(aq) === 3Mg^2+(aq) + 2Au(s)

3. a) Cathode is Co(2) and anode is Pb(2)

E°cell = E°cathode - E°anode = -0.282-(-0.126) = -0.156V

∆G° =-2×96485.3415× -0.156

∆G° is +Ve( non spontaneous)

Similarly Concept and Calculations in case b and c.

If ∆G° is Negative then reaction will be spontaneous.