Electrochemistry 2 Question 8
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8. In the cell, $\mathrm{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $)|\mathrm{HCl}(a q)| \mathrm{AgCl}(s)|\operatorname{Ag}(s)| \operatorname{Pt}(s)$ the cell potential is $0.92 \mathrm{~V}$ when a $10^{-6}$ molal $\mathrm{HCl}$ solution is used. The standard electrode potential of $\left(\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl}^{-}\right)$ electrode is $\left{\right.$ Given, $\frac{2.303 R T}{F}=0.06 \mathrm{~V}$ at $\left.298 \mathrm{~K}\right}$
======= ####8. In the cell, $\mathrm{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $)|\mathrm{HCl}(a q)| \mathrm{AgCl}(s)|\operatorname{Ag}(s)| \operatorname{Pt}(s)$ the cell potential is $0.92 \mathrm{~V}$ when a $10^{-6}$ molal $\mathrm{HCl}$ solution is used. The standard electrode potential of $\left(\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl}^{-}\right)$ electrode is $\left{\right.$ Given, $\frac{2.303 R T}{F}=0.06 \mathrm{~V}$ at $\left.298 \mathrm{~K}\right}$
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $0.40 \mathrm{~V}$
(b) $0.20 \mathrm{~V}$
(c) $0.94 \mathrm{~V}$
(d) $0.76 \mathrm{~V}$
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Solution:
- It is an electrochemical cell. The overall cell reaction can be written, as
$$ \begin{aligned} & \mathrm{H}_{2}(g)+2 \mathrm{AgCl}(s) \longrightarrow 2 \mathrm{HCl}(a q)+2 \mathrm{Ag}(\mathrm{s}) \ & (1 \text { bar }) \quad\left(10^{-6} \mathrm{M}\right) \end{aligned} $$
(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)
$$ \begin{array}{rlrl} \Rightarrow & 0.92 & =\left(E_{\mathrm{c}}^{\circ}-0\right)-0.06 \times \log \frac{\left(10^{-6}\right)^{2} \times 1^{2}}{1 \times 1^{2}} \ & = & E_{\mathrm{c}}^{\circ}+0.06 \times 6 \times 2 \ \Rightarrow & E_{\mathrm{c}}^{\circ}=0.92-0.72=0.20 \mathrm{~V} \end{array} $$
Note $10^{-6}$ molal $\mathrm{HCl}$ is a very dilute solution. So, $10^{-6} \mathrm{~m} \simeq 10^{-6} \mathrm{M}$
