If hydrogen electrode were dipped in a solution of pH =2 at 25° C and 1 bar pressure, the reduction potential of the half-cell would be V.

Question

If hydrogen electrode were dipped in a solution of pH =2 at 25° C and 1 bar pressure, the reduction potential of the half-cell would be V. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Reduction potential of hydrogen electrode, E H =-(2.303 RT / nF ) log (1/[ H +]) (∵ log (1/[ H +])= pH ) =-0.0591 pH =-0.0591 × 2=-0.12 V

Q. If hydrogen electrode were dipped in a solution of $p H = 2$ at $2 5^{\circ} C$ and 1 bar pressure, the reduction potential of the half-cell would be _______ $V$ .

Reduction potential of hydrogen electrode,
$E_{H} = - \frac{2.303 RT}{n F} lo g \frac{1}{[ H ^{+} ]}$
$(∵ lo g \frac{1}{[ H ^{+} ]} = p H)$
$= - 0.0591 p H = - 0.0591 \times 2 = - 0.12 V$

Q. If hydrogen electrode were dipped in a solution of pH=2 at 25∘C and 1 bar pressure, the reduction potential of the half-cell would be _______ V.

Reduction potential of hydrogen electrode, EH​=−nF2.303RT​log[H+]1​ (∵log[H+]1​=pH) =−0.0591pH=−0.0591×2=−0.12V

Q. If hydrogen electrode were dipped in a solution of $p H = 2$ at $2 5^{\circ} C$ and 1 bar pressure, the reduction potential of the half-cell would be _______ $V$ .

Reduction potential of hydrogen electrode,
$E_{H} = - \frac{2.303 RT}{n F} lo g \frac{1}{[ H ^{+} ]}$
$(∵ lo g \frac{1}{[ H ^{+} ]} = p H)$
$= - 0.0591 p H = - 0.0591 \times 2 = - 0.12 V$