Question

Find the emf of the cell reaction: \(\mathrm{Zn(s)+2Ag^+(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}\).
\(\,E^\circ_{\mathrm{Zn^{2+}/Zn}}=-0.76\,\text{V}, E^\circ_{\mathrm{Ag^+/Ag}}=+0.80\,\text{V}\).

Hint:

Always identify cathode (reduction, more positive potential) and anode (oxidation). Use the formula: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] This ensures correct sign and value for the emf of the cell.

Answer 1

Cathode (reduction): \(\mathrm{Ag^+ + e^- \rightarrow Ag}\), \(E^\circ_\text{cathode}=+0.80\,\text{V}\).
Anode (oxidation): \(\mathrm{Zn \rightarrow Zn^{2+}+2e^-}\) (reverse of the reduction couple), \(E^\circ_\text{anode (red)}=-0.76\,\text{V}\).
Standard emf: \(E^\circ_{\text{cell}} = E^\circ_\text{cathode} - E^\circ_\text{anode (red)}\).
\[ E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56\ \text{V} \] \[ \boxed{E^\circ_{\text{cell}} = 1.56\ \text{V}} \]