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Chemistry Question on Electrochemistry

The potential of a hydrogen electrode at pH=10pH = 10 is

A

−0.59 V- 0.59\, V

B

0.00 V0.00\, V

C

0.59 V0.59\, V

D

−0.059-0.059

Answer

−0.59 V- 0.59\, V

Explanation

Solution

The correct answer is Option A) −0.59 V- 0.59\, V

H+(pH=10)∣H2(1H^{+}(p H=10) \mid H_{2}(1 atm )∣Pt(s)) \mid P t(s)
Reaction : 2H+(pH=10)+2e−→H2(12 H^{+}\left(p^{H}=10\right)+2 e^- \rightarrow H_{2}(1 atm ))
E=E0−0.05912log⁡(pH2[H+]2)E=E^{0}-\frac{0.0591}{2} \log \left(\frac{p_{H_{2}}}{\left[H^{+}\right]^{2}}\right)
=0−0.05912log⁡1(10−10)2=0-\frac{0.0591}{2} \log \frac{1}{\left(10^{-10}\right)^{2}}
=−0.05912×2log⁡110−10=-\frac{0.0591}{2} \times 2 \log \frac{1}{10^{-10}}
=−0.0591×10=−0.591=-0.0591 \times 10=-0.591
i.e. E=−0.591 VE=-0.591\, V

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