A galvanic cell is constructed as: Cu(s) | $C u^{2 +}$ (0.5\,M) || $A g^{+}$ (1.0\,M) | Ag(s). Standard reduction potentials are $E_{C u^{2 +} / C u}^{\circ}$ =+0.34\,V and $E_{A g^{+} / A g}^{\circ}$ =+0.80\,V. Calculate the cell emf at 25°C.

easy 2 marks Class 12 · Chemistry · Electrochemistry

Answer

E = 0.47 V

The correct answer is E = 0.47 V — Substitute values: E=0.46-(0.0257/2)·ln(0.5)=0.46-0.01285·(-0.6931)=0.46+0.0089≈0.47 V.

Step-by-step solution

Write the half‑reactions: Cu(s) → $C u^{2 +}$ +2e^- (oxidation) and $A g^{+}$ +e^- → Ag(s) (reduction).
Identify the number of electrons transferred in the overall reaction (n=2).
Calculate the standard cell potential: $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ =0.80-0.34=0.46 V.
Write the reaction quotient Q=[ $C u^{2 +}$ ]/[ $A g^{+}$ ]^{2}=0.5/(1.0)^{2}=0.5.
Use the Nernst equation: E=E°-(0.0257/n)·ln Q.
Substitute values: E=0.46-(0.0257/2)·ln(0.5)=0.46-0.01285·(-0.6931)=0.46+0.0089≈0.47 V.

Common mistake: Using ln Q instead of -ln Q or forgetting the factor n in the denominator

Concept tested: Nernst equation for cell potential

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked