# 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 = 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

cell = 1.1037 V

Step 3: Calculate Ecell using the Nernst Equation. ###### 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