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Chemistry Question on Partial pressure

The pressure of H2H_2 required to make the potential of H2H_2-electrode zero in pure water at 298 K is

A

10−1210^{-12} atm

B

10−1010^{-10} atm

C

10−410^{-4} atm

D

10−1410^{-14} atm

Answer

10−1410^{-14} atm

Explanation

Solution

2H++2e−−>H2(g){2H^{+} + 2e^{-} -> H_2(g)}
EH+/H2=−0.05912log⁡PH2[H+]2E_{H^{+}/H_{2}} = - \frac{0.0591}{2} \log \frac{P_{H_2}}{\left[H^{+}\right]^{2}}
log⁡PH2[H+]2=0,PH2[H+]2=100=1\log \frac{P_{H_2}}{\left[H^{+}\right]^{2}} = 0, \frac{P_{H_2}}{\left[H^{+}\right]^{2}} = 10^{0} = 1
PH2=[H+]2PH_{2} = \left[H^{+}\right]^{2}
For pure H2O;H+=10−7M H_{2}O; H^{+} = 10^{-7} M
PH2=(10−7)2=10−14P_{H_2} = \left(10^{-7}\right)^{2} = 10^{-14} atm