Calculate the emf of the galvanic cell: Cu(s) | $C{u}^{2+}$ (0.01,M) || $Z{n}^{2+}$ (0.1,M) | Zn(s) at 25°C. Given standard potentials $E^{c}ir{c}_{\mathrm{C}{\mathrm{u}}^{2+}/Cu}=+0.34,\mathrm{V}$ and $E^{c}ir{c}_{\mathrm{Z}{\mathrm{n}}^{2+}/Zn}=-0.76,\mathrm{V}$.

Step-by-step solution

1. Determine which half‑cell is the cathode; higher reduction potential is Cu, so Cu(s) | $C{u}^{2+}$ is cathode and Zn(s) | $Z{n}^{2+}$ is anode.
2. Compute the standard cell potential: $E^{c}ir{c}_{\mathrm{cell}}=E^{c}ir{c}_{\mathrm{cathode}}-E^{c}ir{c}_{\mathrm{anode}}=0.34-(-0.76)=1.10,\mathrm{V}$.
3. Write the overall cell reaction: $\mathrm{Zn}(\mathrm{s})+\mathrm{C}{\mathrm{u}}^{2+}(0.01,\mathrm{M})\to \mathrm{Z}{\mathrm{n}}^{2+}(0.1,\mathrm{M})+\mathrm{Cu}(\mathrm{s})$.
4. Calculate the reaction quotient: $Q=\frac{[\mathrm{Z}{\mathrm{n}}^{2+}]}{[\mathrm{C}{\mathrm{u}}^{2+}]}=\frac{0.1}{0.01}=10$. Compute $lnQ=2.3026$. Using Nernst: $E=E_{\mathrm{cell}}^{\circ }-\frac{RT}{nF}lnQ$ with $n=2$, $RT/F=0.0257,\mathrm{V}$. Thus $E=1.10-\frac{0.0257}{2}\times 2.3026=1.10-0.0296=1.0704,\mathrm{V}$. Round to two decimal places to get $1.07,\mathrm{V}$.

Concept tested: Electrochemical cell potential calculation

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked