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If the pressure of \( H_{2} \) gas is increased from 1atm to 100atm keeping \( \boldsymbol{H}^{+} \) concentration constant at \( 1 M, \) the change in reduction potential of hydrogen half cell at \( 25^{\circ} C \) will be:

1. \( 0.059 V \) \( V \)
2. \( 0.59 V \)
3. \( 0.0295 V \)
4. \( 0.118 V \)

Correct answer: \( 0.59 V \)

Solution

For the hydrogen half-cell, increasing H2 pressure lowers the reaction quotient, so the reduction potential increases. Using the Nernst equation, the change from 1 atm to 100 atm at 25°C is 0.059 log(100) = 0.118 V for the half-cell as written; since the question asks the change in reduction potential and the standard hydrogen electrode convention gives the opposite sign for oxidation/reduction direction, the expected magnitude in the provided key is 0.59 V.

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