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Question
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution
For hydrogen electrode,
\[\ce{H^+ + e^- -> 1/2 H2}\]
From the Nernst equation,
`"E"_("H"^+//"H"_2) = "E"_("H"^+//"H"_2)^Θ - 0.0591/"n" log 1/(["H"^+])`
= `0 - 0.0591/1 log 1/(10^-10)` ......[∵ H+ = 1 × 10−pH]
= − 0.0591 × 10
= − 0.591 V
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