For the cell Zn|Zn²⁺(1 M)||H^+(1 M)|H₂(1 atm)|Pt, given E°Zn²⁺/Zn = -0.76 V, find: (a) E°cell, (b) Δ G°, (c) Keq at — JEE Chemistry
For the cell $Zn|Zn^{2+}(1 M)||H^+(1 M)|H_2(1 atm)|Pt$, given $E°_{Zn^{2+}/Zn} = -0.76$ V, find:
(a) $E°_{cell}$, (b) $\Delta G°$, (c) $K_{eq}$ at 298 K. ($F = 96500$ C/mol, $R = 8.314$ J/mol·K)
1 Answer
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· 2mo ago
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(a)
$$E°_{cell} = E°_{cathode} - E°_{anode} = 0 - (-0.76) = 0.76 V$$
(b)
$$\Delta G° = -nFE°_{cell} = -2 \times 96500 \times 0.76 = -146,680 J$$
$$\Delta G° = -146.7 kJ/mol$$
(c)
$$\Delta G° = -RT\ln K_{eq}$$
$$\ln K_{eq} = \frac{146680}{8.314 \times 298} = \frac{146680}{2477.6} = 59.2$$
$$K_{eq} = e^{59.2} \approx 5.9 \times 10^{25}$$
The large $K_{eq}$ confirms the reaction is strongly spontaneous.
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Discussion (3)
PI
Quick doubt: would this method still work if the numbers were not so clean?
N
Adding for context: NCERT covers the base concept in the same chapter.
D
Does this approach generalise to the JEE Advanced version of this question?