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Question: Given that \(E_{H_{2}O|H_{2}|Pt}\)= 0 at 298 K. The pressure of H<sub>2</sub>(g) would be –...

Given that EH2O∣H2∣PtE_{H_{2}O|H_{2}|Pt}= 0 at 298 K. The pressure of H2(g) would be –

A

10–7 bar

B

10–10bar

C

10–12 bar

D

10–14 bar

Answer

10–14 bar

Explanation

Solution

The cell reaction is : 2H+ + 2e → H2

Ecell = Eŗcell – 0.0592\frac{0.059}{2} log pH2[H+]2\frac{p_{H_{2}}}{\lbrack H^{+}\rbrack^{2}}

Eŗcell = 0, Ecell = 0

∴ pH2p_{H_{2}}= [H+]2

In pure water, [H+] = 10–7 (M)

∴ pH2p_{H_{2}} = 10–14 bar.