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In the cell

$Pt_{\left ( s \right )}\mid H_{2}\left ( g, 1 bar \right )\mid HCl_{\left ( aq \right )}\mid AgCl_{\left ( s \right )}\mid Pt_{\left ( s \right )}$ the cell potential is 0.92V when a 10^-6 molal $HCl$ solution is used . The stansard electrode potential of $\left ( AgCl/ag,Cl^{-} \right )$ electrode is: $\left \{ Given \frac{2.303RT}{F}=0.06 V \, at 289 K \right \}$
Option 1)
0.20 V
Option 2)
0.40V
Option 3)
0.76V
Option 4)
0.94V

Answers (1)

best_answer

Nernst Equation -

$M^{n+}(aq)+ne^{-}\rightarrow M(s)$

- wherein

$E_{(M^{n+}/M)}=E_{(M^{n+}/M)}^{0}-\frac{RT}{nf}ln\frac{[M]}{[M^{n+}]}$

$[M^{n+}]$ is concentration of species

F= 96487 C $moi^{-1}$

R= 8.314 $JK^{-1}moi^{-1}$

T= Temperature in kelvin

As we learned in cell potential

At Anode : $H_{2} \rightarrow 2H^{+} + 2 e \times 1$

At Cathode : $e^{-} + AgCl (s) \rightarrow Ag (s) + Cl^{-} (aq)$

$H_{2} + AgCl \rightarrow 2 H^{+} + 2Ag (s) + 2Cl^{-}$

$E_{cell} = E^{o} _{cell} - \frac{0.06}{2}log_{10} (H^{+})^{2} (Cl^{-})^{2}$

$0.925 = \left ( E^{o}_{\frac{H_{2}}{H^{+}}} + E^{o}_{\frac{AgCl}{Ag,Cl^{-}}} \right ) - \frac{0.06}{2} log_{10}\left ((10^{-6})^{2} (10^{-6})^{2} \right )$

$0.92 = 0 + E^{o}_{\frac{AgCl}{Ag,Cl^{-}}} - 0.03 log_{10}(10^{-6})^{4}$

$E^{o}_{\frac{AgCl}{Ag,Cl^{-}}} = 0.92 + 0.3 \times -24$

= 0.2 V

Option 1)
0.20 V
Option 2)

0.40V

Option 3)

0.76V

Option 4)

0.94V

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