Electrochemistry Question 1
Question 1 - 2024 (01 Feb Shift 1)
The potential for the given half cell at $298 \mathrm{~K}$ is
$(-) \ldots \ldots \ldots \ldots \times 10^{-2} \mathrm{~V}$.
$2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{~g})$
$\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}{\mathrm{H}{2}}=2 \mathrm{~atm}$
(Given: 2.303RT $/ \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )
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Answer (1)
Solution
$\mathrm{E}=\mathrm{E}{\mathrm{H}^{+} / \mathrm{H}{2}}^{\circ}-\frac{0.06}{2} \log \frac{\mathrm{P}{\mathrm{H}{2}}}{\left[\mathrm{H}^{+}\right]^{2}}$
$\mathrm{E}=0.00-\frac{0.06}{2} \log \frac{2}{[1]^{2}}$
$\mathrm{E}=-0.03 \times 0.3=-0.9 \times 10^{-2} \mathrm{~V}$
