The standard electrode potential of the Ni2+/Ni electrode is −0.236 V. This electrode is combined with a hydrogen electrode (H2 at 1 bar) dipping in an acid solution. At what pH of the acid solution will the measured emf of the cell be zero at 25°C? (Take [Ni2+] = 1 M and 2.303RT/F = 0.059 V)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
Nickel has the lower standard electrode potential, so it is the anode (left) and the hydrogen electrode is the cathode (right):
Anode: \( \displaystyle \ce{Ni(s) -> Ni^2+(aq) + 2e^-} \)
Cathode: \( \displaystyle \ce{2H^+(aq) + 2e^- -> H2(g)} \)
Cell reaction: \( \displaystyle \ce{Ni(s) + 2H^+(aq) -> Ni^2+(aq) + H2(g)} \qquad n=2 \)
Standard cell potential:
\( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}_{\text{right}}-E^{\circ}_{\text{left}}=0-(-0.236)=0.236\,\mathrm{V} \)
Nernst equation (pH2 = 1 bar):
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{2}\log\frac{[\ce{Ni^2+}]}{[\ce{H^+}]^2} \)
Substituting [\( \ce{Ni^2+} \)] = 1 M, with \( \log\frac{1}{[\ce{H^+}]^2}=-2\log[\ce{H^+}]=2\,\mathrm{pH} \):
\( \displaystyle E_{\text{cell}}=0.236-0.059\,\mathrm{pH} \)
Setting Ecell = 0:
\( \displaystyle \mathrm{pH}=\frac{0.236}{0.059}=4 \)