Question

To find the standard potential of \(\mathrm{M^{3+}/M}\) electrode, the following cell is constituted :
\(\mathrm{Pt/M/M^{3+}(0.001\ mol\ L^{-1})/Ag^{+}(0.01\ mol\ L^{-1})/Ag}\)
The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction \(\mathrm{M^{3+} + 3e^{-} \to M}\) at 298 K will be :
(Given \(E^{\ominus}_{\mathrm{Ag^{+}/Ag}}\) at 298 K = 0.80 Volt)

• A. 0.38 Volt
• B. 0.32 Volt
• C. 1.28 Volt
• D. 0.66 Volt

Hint:

Write the cell reaction \(\mathrm{M + 3Ag^+ \to M^{3+} + 3Ag}\) (n = 3), apply the Nernst equation to get \(E^\ominus_{cell}\), then subtract from the silver standard potential.

Answer 1

The Correct Option is B

Anode: \(\mathrm{M \to M^{3+} + 3e^-}\); cathode: \(\mathrm{Ag^+ + e^- \to Ag}\). Cell reaction: \(\mathrm{M + 3Ag^+ \to M^{3+} + 3Ag}\), n = 3.
Nernst: \(E_{cell} = E^\ominus_{cell} - \frac{0.059}{3}\log\frac{[\mathrm{M^{3+}}]}{[\mathrm{Ag^+}]^3} = E^\ominus_{cell} - \frac{0.059}{3}\log\frac{10^{-3}}{(10^{-2})^3}\).
\(\frac{10^{-3}}{10^{-6}} = 10^{3}\), so \(\log = 3\) and \(E_{cell} = E^\ominus_{cell} - 0.059\).
\(E^\ominus_{cell} = 0.421 + 0.059 = 0.48\) V.
\(E^\ominus_{cell} = E^\ominus_{\mathrm{Ag^+/Ag}} - E^\ominus_{\mathrm{M^{3+}/M}}\), so \(E^\ominus_{\mathrm{M^{3+}/M}} = 0.80 - 0.48 = 0.32\) V.
Answer: (2) 0.32 Volt