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A concentration cell is constructed using two \mathrm{Ag} / \mathrm{AgCl} electrodes. The initial concentration of  \mathrm{Cl^{-}}in the anode compartment is \mathrm{0.001M}  and that in the cathode compartment is \mathrm{0.1M} The cell is allowed to come to equilibrium. Calculate the cell potential at \mathrm{298K.} Given, \mathrm{E_{\mathrm{Ag} / \mathrm{AgCl}}^{\circ}=+0.222 \mathrm{ \ V \ and\ R}=8.314 \mathrm{~J} / \mathrm{K} . \mathrm{mol}}

 

Option: 1

0.020 V


Option: 2

0.060 V


Option: 3

0.080 V


Option: 4

0.100 V


Answers (1)

best_answer

The overall reaction in the cell can be represented as:

\begin{aligned} & A g(s)+C l^{-}(0.1 M) \rightarrow A g C l(s)+e^{-} \\ & A g C l(s)+e^{-} \rightarrow A g(s)+C l^{-}(0.001 M) \end{aligned}

The cell reaction is : A g(s)+C l^{-}(0.1 M) \rightarrow A g(s)+C l^{-}(0.001 M)

The Nernst equation for the cell can be written as: E=E^{\circ}-\frac{R T}{n F} \ln Q

where E^{\circ}cell is the standard cell potential, R is the gas constant, T is the temperature in kelvins,n is the number of electrons transferred in the cell reaction,F is the Faraday constant, and Q is the reaction quotient.

Since the cell is at equilibrium, Q=K (the equilibrium constant) and \Delta G=0 Using the equation \Delta G=-n F E and solving for E, we get:

E=\frac{\Delta G}{n F}=\frac{R T \ln K}{n F}

The equilibrium constant for the cell reaction can be calculated as:

K=\frac{[\mathrm{AgCl}]\left[C l^{-}\right]}{[\mathrm{Ag}]\left[C l^{-}\right]^2}

At equilibrium,

\\\ [\mathrm{AgCl}]=K_{s p}=1.8 \times 10^{-10}(\text { at } 298 \mathrm{~K}), \\ \\\mathrm{\left[\mathrm{Cl}^{-}\right] \text {anode }=0.001 \mathrm{M}, \text { and }\left[\mathrm{Cl}^{-}\right] \text {cathode }=0.1 M}

Substituting the values in the equation for E , we get:

E=0.222-\frac{0.0257}{1} \ln \frac{0.1 \times 1.8 \times 10^{-10}}{(0.001)^2}

Solving for E, we get E = 0.080 V.

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Gunjita

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