(A) For a concentration cell consisting of two hydrogen electrodes,the $\text{EMF}$ is given by the Nernst equation: $E_{\text{cell}} = -0.059 \log \f

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(A) For a concentration cell consisting of two hydrogen electrodes,the $\text{EMF}$ is given by the Nernst equation: $E_{\text{cell}} = -0.059 \log \frac{[H^+]_1}{[H^+]_2}$ Given $E_{\text{cell}} = 0.17 \ V$ and $[H^+]_2 = 10^{-3} \ M$: $0.17 = -0.059 \log \frac{[H^+]_1}{10^{-3}}$ $-2.88 = \log [H^+]_1 - \log(10^{-3})$ $-2.88 = \log [H^+]_1 - (-3)$ $-2.88 = \log [H^+]_1 + 3$ $-\log [H^+]_1 = 3 + 2.88 = 5.88$ Since $\text{pH} = -\log [H^+]$,the $\text{pH}$ at the other electrode is $5.88$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is:

TS EAMCET 2018 All TS EAMCET

A
$5.88$
B
$4.88$
C
$2.08$
D
$3.08$

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The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is:

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