Question

For the cell \(\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,||\,M^{x+}(aq)\,|\,M(s)}\), different half cells and their standard electrode potentials are given below :

\(\mathrm{M^{x+}(aq)/M(s)}\)	\(\mathrm{Au^{3+}(aq)/Au(s)}\)	\(\mathrm{Ag^{+}(aq)/Ag(s)}\)	\(\mathrm{Fe^{3+}(aq)/Fe^{2+}(aq)}\)	\(\mathrm{Fe^{2+}(aq)/Fe(s)}\)
\(E^\circ_{M^{x+}/M}\) (V)	1.40	0.80	0.77	\(-0.44\)

If \(E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76\) V, which cathode will give maximum value of \(E^\circ_{cell}\) per electron transferred?

• A. \(\mathrm{Ag^{+}/Ag}\)
• B. \(\mathrm{Fe^{3+}/Fe^{2+}}\)
• C. \(\mathrm{Au^{3+}/Au}\)
• D. \(\mathrm{Fe^{2+}/Fe}\)

Hint:

\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\) with Zn as the anode; divide each \(E^\circ_{cell}\) by the number of electrons n transferred (3 for Au, 1 for Ag, 1 for Fe3+/Fe2+, 2 for Fe2+/Fe).

Answer 1

The Correct Option is A

Anode is Zn: \(E^\circ_{cell} = E^\circ_{cathode} - (-0.76) = E^\circ_{cathode} + 0.76\) V.
\(\mathrm{Au^{3+}/Au}\): \(E^\circ_{cell} = 1.40 + 0.76 = 2.16\) V, n = 3, so per electron \(= 2.16/3 = 0.72\) V.
\(\mathrm{Ag^{+}/Ag}\): \(E^\circ_{cell} = 0.80 + 0.76 = 1.56\) V, n = 1, so per electron \(= 1.56\) V.
\(\mathrm{Fe^{3+}/Fe^{2+}}\): \(E^\circ_{cell} = 0.77 + 0.76 = 1.53\) V, n = 1, so per electron \(= 1.53\) V.
\(\mathrm{Fe^{2+}/Fe}\): \(E^\circ_{cell} = -0.44 + 0.76 = 0.32\) V, n = 2, so per electron \(= 0.16\) V.
The maximum value per electron is 1.56 V for \(\mathrm{Ag^{+}/Ag}\).
Answer: (1) \(\mathrm{Ag^{+}/Ag}\)