The electrochemical cell is set up as follows | vedclass

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(A) The cell is a concentration cell where $E_{cell} = \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
For $HCl$,which is a strong acid,the concentration of

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The $pH$ of $0.1 \, M \, HCl$ and $0.1 \, M$ acetic acid are not equal.

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(A) The cell is a concentration cell where $E_{cell} = \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
For $HCl$,which is a strong acid,the concentration of $H^{+}$ ions is $C_1 = 0.1 \, M$.
For acetic acid $(CH_3COOH)$,which is a weak acid,the concentration of $H^{+}$ ions is $C_2 = 0.1 	imes \alpha \, M$,where $\alpha$ is the degree of dissociation.
Since $\alpha < 1$,$C_2 
eq C_1$.
Therefore,$E_{cell} = 0.0591 \log \frac{0.1 \alpha}{0.1} = 0.0591 \log \alpha$.
Since $\alpha 
eq 1$,$E_{cell} 
eq 0$.

The electrochemical cell is set up as follows: $Pt (H_2, 1 \, atm) | HCl (0.1 \, M) || \text{Acetic Acid} (0.1 \, M) | (H_2, 1 \, atm) Pt$. The $E.M.F.$ of the cell is not zero because:

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