Calculate the emf of the cell at 25 °C: Zn(s) | $Z{n}^{2+}$ (0.01 M) || $C{u}^{2+}$ (1.0 M) | Cu(s). Standard potentials: $E_{Z{n}^{2+}/Zn}^{\circ }$= -0.76 V, $E_{C{u}^{2+}/Cu}^{\circ }$= +0.34 V. Use the Nernst equation.

Step-by-step solution

1. Write the overall cell reaction: Zn(s) + $C{u}^{2+}\to Z{n}^{2+}$ + Cu(s).
2. Identify that the number of electrons transferred, n, is 2.
3. Calculate the standard cell potential: E°_cell = $E_{cathode}^{\circ }-E_{anode}^{\circ }$ = 0.34 - (-0.76) = 1.10 V.
4. Apply the Nernst equation: E = E° - (0.0592/n) log([$Z{n}^{2+}$]/[$C{u}^{2+}$]).
5. Insert the concentrations: log([$Z{n}^{2+}$]/[$C{u}^{2+}$]) = log(0.01/1) = log(0.01) = -2.
6. Compute the correction term: (0.0592/2)(-2) = -0.0592 V, so E = 1.10 - (-0.0592) = 1.1592 V.
7. Round the result to two decimal places.

Concept tested: Nernst equation for cell potential

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked