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Solve the following.
Calculate the value of Δ G° for the following cell at 25°C. Zn(s) / Zn²⁺(1M) ∥ Sn²⁺(1M) / Sn(s)…
Calculate the value of $\Delta G^\circ$ for the following cell at $25^\circ\text{C}$.
$\text{Zn(s)} / \text{Zn}^{2+}(1\text{M}) \parallel \text{Sn}^{2+}(1\text{M}) / \text{Sn(s)}$
Given $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0\cdot76\text{ V}, E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0\cdot14\text{ V}$
$1\text{ faraday} = 96,500\text{ coulombs}$
Answer
Answer
Formula:
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}}$
$\Delta G^\circ = -n F E^\circ_{\text{cell}}$
Substitution:
$E^\circ_{\text{cell}} = -0\cdot14\text{ V} - (-0\cdot76\text{ V}) = +0\cdot62\text{ V}$
Here, $n = 2$ electrons are transferred.
$\Delta G^\circ = -2 \times 96,500\text{ C mol}^{-1} \times 0\cdot62\text{ V}$
$= -119,660\text{ J mol}^{-1} = -119\cdot66\text{ kJ mol}^{-1}$
Final answer: -119.66 $\text{kJ mol}^{-1}$