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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_________ \(\times 10^{-2} \mathrm{~V}\).\(\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)\)

  1. A \(30\)
  2. B \(75\)
  3. C \(45\)
  4. D \(18\)
Verified Solution

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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