Problem: What will be the cell potential for a Cu−Zn galvanic cell with [Zn2+] = 0.1 M and the [Cu2+] = 0.01 M?a) 1.10 Vb) 1.13 Vc) 1.04 Vd) 1.07 V

FREE Expert Solution

Nernst Equation

Ecell=E°cell-(0.05916 Vn) log Q

Ecell = cell potential under non-standard conditions
cell = standard cell potential
n = mole e- transferred
Q = reaction quotient = products/reactants



Step 1: Identify the anode and the cathode in the reaction and write the overall reaction

• ↓ E° → oxidation → anode
• ↑ E° → reduction → cathode


Cu2+ + 2 e- → Cu                    E° = 0.3419 V → cathode
Zn2+ + 2 e- → Zn                     E° = 0.7618 V anode


lose electrons → oxidation → anode
gain electrons → reduction → cathode


Balance the electrons:

Cathode:  Cu2+ + 2 e- → Cu
Anode:                   Zn → Zn2+ + 2 e- 

Overall reaction: Cu2+ + Zn → Cu + Zn2+ 

# of electrons transferred = 2 e-



Step 2: Calculate the cell potential of the reaction.

Cu2+ + 2 e- → Cu                    E° = 0.3419 V → cathode
Zn2+ + 2 e- → Zn                     E° = 0.7618 V anode


E°cell=E°cathode-E°anodeE°cell=0.3419 V-(-0.7618 V)E°cell=0.3419 V+0.7618 V

cell = 1.1037 V


Step 3: Calculate Ecell using the Nernst Equation.

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Problem Details

What will be the cell potential for a Cu−Zn galvanic cell with [Zn2+] = 0.1 M and the [Cu2+] = 0.01 M?

a) 1.10 V

b) 1.13 V

c) 1.04 V

d) 1.07 V

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