The emf of the cell M|Mn+(0.02 M)||H+(1 M)|H2(g, 1 bar)|Pt at 25°C is 0.81 V. The standard electrode potential of the Mn+/M electrode is −0.76 V. What is the charge n on the metal ion? (2.303RT/F = 0.059 V at 298 K)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The metal M is the anode (left) and the hydrogen electrode is the cathode (right). The cell reaction and its Nernst equation are:
\( \displaystyle \ce{M(s)}+n\,\ce{H^+(aq)}\rightarrow\ce{M^{n+}(aq)}+\frac{n}{2}\,\ce{H2(g)} \)
\( \displaystyle E_{(cell)}=E^{\circ}_{(cell)}-\frac{0.059}{n}\log\frac{[\ce{M^{n+}}]}{[\ce{H^+}]^{n}} \)
Standard emf:
\( \displaystyle E^{\circ}_{(cell)}=E^{\circ}_{right}-E^{\circ}_{left}=0-(-0.76\,\mathrm{V})=0.76\,\mathrm{V} \)
Substitution ([\( \ce{M^{n+}} \)] = 0.02 M, [\( \ce{H^+} \)] = 1 M, \( \ce{H2} \) at 1 bar):
\( \displaystyle 0.81=0.76-\frac{0.059}{n}\log0.02 \)
Using log 0.02 = −1.70:
\( \displaystyle 0.81-0.76=\frac{0.059\times1.70}{n} \)
\( \displaystyle n=\frac{0.100}{0.05}=2 \)
The metal ion carries a charge of +2 (\( \ce{M^2+} \)).