Electrochemistry Result Question 8
8. In the cell, $\mathrm{Pt}(s)\left|\mathrm{H} _2(g, 1 \mathrm{bar})\right| \mathrm{HCl}(a q)|\mathrm{AgCl}(s)| \mathrm{Ag}(s) \mid \mathrm{Pt}(s)$ the cell potential is $0.92 \mathrm{~V}$ when a $10^{-6}$ molal $\mathrm{HCl}$ solution is used. The standard electrode potential of $(\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl})$ electrode is Given, $\frac{2.303 R T}{F} 0.06 \mathrm{~V}$ at $298 \mathrm{~K}$
(a) $0.40 \mathrm{~V}$
(b) $0.20 \mathrm{~V}$
(c) $0.94 \mathrm{~V}$
(d) $0.76 \mathrm{~V}$
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Answer:
Correct Answer: 8. ( b )
Solution:
- It is an electrochemical cell. The overall cell reaction can be written, as
$ \underset{(1 \text { bar })}{\mathrm{H}_2(g)}+2 \mathrm{AgCl}(s) \longrightarrow \underset{\left(10^{-6} \mathrm{M}\right)}{2 \mathrm{HCl}(a q)}+2 \mathrm{Ag}(\mathrm{s}) $
(i) According to Nernst equation,
$ E_{\text {cell }}=\left(E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\right)-\frac{2.303 \times R T}{n \times F} \log \frac{[\mathrm{HCl}]^2[\mathrm{Ag}]^2}{p_{\mathrm{H}_2}[\mathrm{AgCl}]^2} $
Here, (i) $E_{\mathrm{c}}^{\circ}=E_{\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl}^{-}}^{\circ}=E_{\text {cathode }}^{\circ}$
(ii) $E_{\text {anode }}^{\circ}=E_{2 \mathrm{H}^{+} / \mathrm{H}_2}^{\circ}=0.00 \mathrm{~V}$ (Standard hydrogen electrode)
$ \begin{aligned} \Rightarrow \quad 0.92 & =\left(E_{\mathrm{c}}^{\circ}-0\right)-0.06 \times \log \frac{\left(10^{-6}\right)^2 \times 1^2}{1 \times 1^2} \\ & =E_{\mathrm{c}}^{\circ}+0.06 \times 6 \times 2 \\ \Rightarrow \quad & E_{\mathrm{c}}^{\circ}=0.92-0.72=0.20 \mathrm{~V} \end{aligned} $
Note: $10^{-6}$ molal HCl is a very dilute solution.
So, $10^{-6} \mathrm{~m} \simeq 10^{-6} \mathrm{M}$
BETA
