Question:
Calculate the emf of the following cell at 25$^\circ$C :
Zn(s) $\mid$ Zn$^{2+}$ (0.1 M) $\parallel$ H$^+$ (0.01 M) $\mid$ H$_2$(g) (1 bar), Pt(s)
Given: E$^\circ_{Zn^{2+}/Zn}$ = –0.76 V, E$^\circ_{2H^+/H_2}$ = 0.00 V, log 10 = 1
Calculate the emf of the following cell at 25$^\circ$C :
Zn(s) $\mid$ Zn$^{2+}$ (0.1 M) $\parallel$ H$^+$ (0.01 M) $\mid$ H$_2$(g) (1 bar), Pt(s)
Given: E$^\circ_{Zn^{2+}/Zn}$ = –0.76 V, E$^\circ_{2H^+/H_2}$ = 0.00 V, log 10 = 1
Zn(s) $\mid$ Zn$^{2+}$ (0.1 M) $\parallel$ H$^+$ (0.01 M) $\mid$ H$_2$(g) (1 bar), Pt(s)
Given: E$^\circ_{Zn^{2+}/Zn}$ = –0.76 V, E$^\circ_{2H^+/H_2}$ = 0.00 V, log 10 = 1
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Always identify anode/cathode → use Nernst carefully → plug units right!
Updated On: Jul 3, 2025
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Solution and Explanation
Cell reaction: Zn(s) + 2H$^+$ → Zn$^{2+}$ + H$_2$(g)
E$^\circ_{cell}$ = E$^\circ_{cathode}$ – E$^\circ_{anode}$
= 0.00 V – (–0.76 V) = +0.76 V.
Nernst equation:
E$_{cell}$ = E$^\circ_{cell}$ – $\frac{0.0591}{n}$ log $\frac{[Zn^{2+}]}{[H^+]^2}$
Here, n = 2.
Substitute:
E$_{cell}$ = 0.76 – $\frac{0.0591}{2}$ log $\frac{0.1}{(0.01)^2}$
= 0.76 – 0.02955 × log (1000)
= 0.76 – 0.02955 × 3
= 0.76 – 0.08865
= 0.671 V.
E$^\circ_{cell}$ = E$^\circ_{cathode}$ – E$^\circ_{anode}$
= 0.00 V – (–0.76 V) = +0.76 V.
Nernst equation:
E$_{cell}$ = E$^\circ_{cell}$ – $\frac{0.0591}{n}$ log $\frac{[Zn^{2+}]}{[H^+]^2}$
Here, n = 2.
Substitute:
E$_{cell}$ = 0.76 – $\frac{0.0591}{2}$ log $\frac{0.1}{(0.01)^2}$
= 0.76 – 0.02955 × log (1000)
= 0.76 – 0.02955 × 3
= 0.76 – 0.08865
= 0.671 V.
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