JEE MAIN - Chemistry (2024 - 27th January Evening Shift - No. 25)
The hydrogen electrode is dipped in a solution of $$\mathrm{pH}=3$$ at $$25^{\circ} \mathrm{C}$$. The potential of the electrode will be _________ $$\times 10^{-2} \mathrm{~V}$$.
$$\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$$
Answer
18
Explanation
$$\begin{aligned}
& 2 \mathrm{H}_{(\mathrm{aq} .)}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \\
& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \\
& =0-0.059 \times 3=-0.177 \text { volts. }=-17.7 \times 10^{-2} \mathrm{~V}
\end{aligned}$$
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