The hydrogen electrode is dipped in a solution of pH - 3 at 25- C- The potential of the cell would be the value of 2-303 RT-Fis 0-059 V

Reduction potential of hydrogen electrode =E_H = 2.303 RT/F log 1/[H^+]= 0.059 pH = 0.059 × 3 = 0.177 V ...

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Nernst Equation

The hydrogen ...

Question

The hydrogen electrode is dipped in a solution of $p H = 3$ at $25^{\circ} C$ . The potential of the cell would be (the value of $2.303 \frac{R T}{F}$ is $0.059 V$

0.177 V

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- 0.177 V

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0.087 V

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0.059 V

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