A cell Cu | Cu2+ || Ag+ | Ag was made up using 0.01 M AgNO3 and 0.0001 M Cu(NO3)2 the cell potential at 298 K is
Given \( E^\circ_{\text{cell}} = 0.46 \text{ V}, \ \displaystyle\frac{2.303RT}{F} = 0.059 \text{ V} \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → +0.46 V
The cell reaction is
\( \ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)} \), n = 2
By the Nernst equation:
\( \displaystyle E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2}\log\frac{[\ce{Cu^2+}]}{[\ce{Ag+}]^2} = 0.46 - \frac{0.059}{2}\log\frac{10^{-4}}{(10^{-2})^2} \)
\( \displaystyle = 0.46 - \frac{0.059}{2}\log 1 = 0.46 - 0 = +0.46\ \text{V} \)
The concentration term is 1, so the cell potential equals the standard value.