Electrochemistry Question 6
Question 6 - 2024 (27 Jan Shift 2)
The hydrogen electrode is dipped in a solution of $\mathrm{pH}=3$ at $25^{\circ} \mathrm{C}$. The potential of the electrode will be _______ $\times 10^{-2} \mathrm{~V}$.
$\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$
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Answer (18)
Solution
$2 \mathrm{H}{\text {(aq.) }}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}{2}(\mathrm{~g})$
$\mathrm{E}{\text {cell }}=\mathrm{E}{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{\mathrm{P}{\mathrm{H}{2}}}{\left[\mathrm{H}^{+}\right]^{2}}$
$=0-0.059 \times 3=-0.177$ volts. $=-17.7 \times 10^{-2} \mathrm{~V}$.
