Consider the following electrochemical cell reaction:
Mg (s) + 2Ag+(aq, 0.1 M) → Mg2+ (aq, 0.01 M) + 2Ag (s)
The emf of the cell in Volts is
(E°(cell) = 3.17 V)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3.17
The Nernst equation gives the emf of a cell at any concentration. At 298 K, for a cell reaction with n electrons transferred:
\( \displaystyle E_{\mathrm{cell}}=E^\circ_{\mathrm{cell}}-\frac{0.059}{n}\log Q \)
The reaction quotient Q is written as products over reactants, each concentration raised to its coefficient; solids are left out.
The electrode reactions of the cell are:
\( \displaystyle \text{Anode: }\ce{Mg(s) -> Mg^2+(aq) + 2e^-}\quad\text{Cathode: }\ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \)
Two electrons are transferred, so n = 2:
\( \displaystyle Q=\frac{[\ce{Mg^2+}]}{[\ce{Ag+}]^2}=\frac{0.01}{(0.1)^2}=\frac{0.01}{0.01}=1 \)
\( \displaystyle E_{\mathrm{cell}}=3.17-\frac{0.059}{2}\log 1 \)
\( \displaystyle E_{\mathrm{cell}}=3.17-0=3.17\ \mathrm{V} \)
So the emf of the cell is 3.17 V.
Shortcut: [Ag+]2 = (0.1)2 = 0.01 equals \( \ce{[Mg^{2+}]} \), so Q = 1, log Q = 0 and Ecell = E°cell. No calculation is needed.