Q5 / 90 · JEE Main 2011 · Chemistry

The reduction potential of hydrogen half cell will be negative if :

  • A
    $\mathrm{p}\left(\mathrm{H}_2\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$
  • Correct answer
    B
    $\mathrm{p}\left(\mathrm{H}_2\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$
  • C
    $\mathrm{p}\left(\mathrm{H}_2\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$
  • D
    $\mathrm{p}\left(\mathrm{H}_2\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$

Solution

$2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})$ $\mathrm{E}=\mathrm{E}^{\circ}-0.059 \log \left(\frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\right)$ (here $\mathrm{E}$ is -ve when $\mathrm{P}_{\mathrm{H}_2}>\left[\mathrm{H}^{+}\right]^2$ ) $=\frac{-0.0591}{2} \log _{10}\left(\frac{2}{1}\right)=\frac{-.0591}{2} \times .3010=$ negative value

Hence, the correct answer is option: 'B'.

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