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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

The potential for the given half cell at \(298 \mathrm{~K}\) is \((-)\) _______ \(\times 10^{-2} \mathrm{~V}\). \(2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})\) \({\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}}\) (Given: \(2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3\) )

  1. A \(0\)
  2. B \(1\)
  3. C \(3\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(E=E_{\mathrm{H}^{+} / \mathrm{H}_2}^0-\frac{0.06}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\) \(\mathrm{E}=0.00-\frac{0.06}{2} \log \frac{2}{[1]^2}\) \(E=-0.03 \times 0.3=-0.9 \times 10^{-2} \mathrm{~V}\)