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

Nernst Equation

$\boxed{{\mathbf{E}}_{\mathbf{cell}}\mathbf{=}\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{-}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{05916}\mathbf{ }\mathbf{V}}{\mathbf{n}}\mathbf{\right)}\mathbf{ }\mathbf{log}\mathbf{ }\mathbf{Q}}$

E_cell = cell potential under non-standard conditions
E°_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

$$\begin{gathered} \boxed{\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\mathbf{E}{\mathbf{°}}_{\mathbf{cathode}}\mathbf{-}\mathbf{E}{\mathbf{°}}_{\mathbf{anode}}} \\ \mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3419}\mathbf{ }\mathbf{V}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{7618}\mathbf{ }\mathbf{V}\mathbf{)} \\ \mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3419}\mathbf{ }\mathbf{V}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{7618}\mathbf{ }\mathbf{V} \end{gathered}$$

E°_cell = 1.1037 V

Step 3: Calculate E_cell using the Nernst Equation.