Calculate the emf of the cell $\mathrm{Zn}(\mathrm{s})|\mathrm{Z}{\mathrm{n}}^{2+}(0.001\mathrm{M})||\mathrm{A}{\mathrm{g}}^{+}(0.10\mathrm{M})|\mathrm{Ag}(\mathrm{s})$ at $25^{\circ }\text{C}$. Standard potentials: $E_{\mathrm{Z}{\mathrm{n}}^{2+}/Zn}^{\circ }=-0.76\text{V}$, $E_{\mathrm{A}{\mathrm{g}}^{+}/Ag}^{\circ }=+0.80\text{V}$.

Step-by-step solution

1. Write the overall cell reaction: $\mathrm{Zn}(\mathrm{s})+2A{\mathrm{g}}^{+}(\mathrm{aq})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+2Ag(\mathrm{s})$.
2. Determine the number of electrons transferred, $n=2$.
3. Calculate the standard cell potential: $E^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=0.80-(-0.76)=1.56\text{V}$.
4. Write the reaction quotient: $Q=\frac{[\mathrm{Z}{\mathrm{n}}^{2+}]}{[\mathrm{A}{\mathrm{g}}^{+}{]}^2}=\frac{0.001}{(0.10{)}^2}=0.1$.
5. Apply the Nernst equation: $E=E^{\circ }-\frac{0.0592}{n}\log Q$.
6. Substitute values: $E=1.56-\frac{0.0592}{2}\log 0.1=1.56-0.0296(-1)=1.56+0.0296=1.5896\text{V}$.
7. Round to two decimal places: $E\approx 1.59\text{V}$.

Concept tested: Using Nernst equation for cell emf

Published 5 September 2026 · How this solution was prepared and checked