Question

The reduction potential of hydrogen electrode at ph 10 is

Answer 1

Answer:

-0.0591 V

Explanation:

pH =10

Using Formula [H+] = 10^-pH

So [H+] = 10^-10 M

Electrode reaction is

H+ + e^- ---> 1/2 H2

Use The Formula

Ecell = E°cell -0.0591/n log [H2] / [H+]

E°cell= 0 and n= 1

Ecell = -0.0591 log(1/10^-10)

Ecell = -0.0591 × 10 = -0.0591 V

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