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)
A admin

 

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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