calculate d potential of h2 electrode in contact wid a solution whose ph is 10.

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Solution

Given:
pH = 10
as we know pH = -log [H⁺]
Thus for pH = 10 the H⁺ ions conc. will be : [10 ^-10 ]

Here H⁺ ion accepts 1 electron and gets reduced to H atom

According to the Nernst equation:

$E_{c e l l} = E_{0} - \frac{0.0591}{1} \log \frac{[H_{2}]}{[H^{+}]} H e r e v a l u e o f E_{0} = 0 a s s \tan d a r d v a l u e o f H_{2} g a s = 0 E_{c e l l} = - 0.0591 \log \frac{1}{10^{- 10}} = - 0.0591 x 10 = - 0.591 V$

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