Electrochemistry Question 6
Question 6 - 2024 (27 Jan Shift 2)
The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ} C$. The potential of the electrode will be _______ $\times 10^{-2} V$.
$\left(\frac{2.303 RT}{F}=0.059 V\right)$
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Answer (18)
Solution
$2 H _{\text {(aq.) }}^{+}+2 e^{-} \rightarrow H _2(g)$
$E _{\text {cell }}=E _{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{P _{H _2}}{\left[H^{+}\right]^{2}}$
$=0-0.059 \times 3=-0.177$ volts. $=-17.7 \times 10^{-2} V$.
