JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=3\) at \(25^{\circ} \mathrm{C}\). The potential of the electrode will be_________ \(\times 10^{-2} \mathrm{~V}\).\(\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)\)
- A \(30\)
- B \(75\)
- C \(45\)
- D \(18\)
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
Detailed explanation
\( 2 \mathrm{H}_{\text {(aq.) }}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \) \( \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \)…
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