A cell is written as
\( Mg| Mg^{2+} (0.0001M)|| Ag^+ (0.01M)|Ag \)
Calculate its \( E_{cell} \) if \( E°_{cell} = 1.25 V \) at 298 K
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.25 V
The emf of a cell at concentrations other than 1 M is given by the Nernst equation. First write the electrode reactions. Magnesium, on the left, is the anode and is oxidised; silver ions are reduced at the cathode:
\( \displaystyle \text{Anode: } \ce{Mg(s) -> Mg^2+(aq) + 2e-} \)
\( \displaystyle \text{Cathode: } \ce{2Ag+(aq) + 2e- -> 2Ag(s)} \)
\( \displaystyle \text{Cell: } \ce{Mg(s) + 2Ag+(aq) -> Mg^2+(aq) + 2Ag(s)}, \quad n = 2 \)
At 298 K:
\( \displaystyle E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\log\frac{[\ce{Mg^2+}]}{[\ce{Ag+}]^{2}} \)
\( \displaystyle E_{\text{cell}} = 1.25 - \frac{0.059}{2}\log\frac{10^{-4}}{(10^{-2})^{2}} \)
\( \displaystyle E_{\text{cell}} = 1.25 - 0.0295\log\frac{10^{-4}}{10^{-4}} = 1.25 - 0.0295\log 1 \)
\( \displaystyle E_{\text{cell}} = 1.25 - 0 = 1.25\ \mathrm{V} \)
So the emf of the cell is 1.25 V, the same as the standard value.
Shortcut: \( [\ce{Mg^2+}] = 10^{-4} \) M and \( [\ce{Ag+}]^{2} = (10^{-2})^{2} = 10^{-4} \) M, so the reaction quotient is 1, its log is zero and \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \).