Question

Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10.​

Answer 1

Answer:

here is the answer

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Answer 2

Answer:

• -0.591 V

Explanation:

Given

• pH of solution is 10

Let us take

◉ T = 298 K

◉ F = 96500

◉ R = 8.314

For hydrogen electrode

◉ $H^{+} + e^{-} \:\longrightarrow\:\frac{1}{2}H_{2}$

$[H^{+}] = 10^{-10}M$

Using Nernest equation

$\boxed{ E_{(H^{+}/\frac{1}{2}H_{2})} = E^{0}_{(H^{+}/H_{2})}-\frac{RT}{nF}ln\frac{H_{2}}{[H^{+}]} }\:\:\:\:\:P_{H_{2}} = unity(1)$

$\implies\:\:E^{0}_{(H^{+}/H_{2})}-\frac{0.0591}{1}log\frac{1}{[H^{+}]}$

$\implies\:\:0-\frac{0.0591}{1}log\frac{1}{[10^{-10}]}$

$\implies\:\:-0.0591\:log\:10^{10}$

$\implies\:\: -0.591\: V$