Electrochemistry Question 1
Question 1 - 2024 (01 Feb Shift 1)
The potential for the given half cell at $298 K$ is
$(-) \ldots \ldots \ldots \ldots \times 10^{-2} V$.
$2 H^{+}(aq)+2 e^{-} \rightarrow H _2(g)$
$\left[H^{+}\right]=1 M, P _{H _2}=2 atm$
(Given: 2.303RT $/ F=0.06 V, \log 2=0.3$ )
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Answer (1)
Solution
$E=E _{H^{+} / H _2}^{\circ}-\frac{0.06}{2} \log \frac{P _{H _2}}{\left[H^{+}\right]^{2}}$
$E=0.00-\frac{0.06}{2} \log \frac{2}{[1]^{2}}$
$E=-0.03 \times 0.3=-0.9 \times 10^{-2} V$
