For the cell: Ag | AgCl | KCl(1 M) || KBr(1 | vedclass

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(D) The cell reaction is: $AgCl(s) + Br^-(aq) \rightleftharpoons AgBr(s) + Cl^-(aq)$.Since the concentrations of $KCl$ and $KBr$ are both $1 \ M$,the

Vedclass · Accepted Answer

$E_{cell} 
eq 0$

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(D) The cell reaction is: $AgCl(s) + Br^-(aq) ightleftharpoons AgBr(s) + Cl^-(aq)$.Since the concentrations of $KCl$ and $KBr$ are both $1 \ M$,the reaction quotient $Q = \frac{[Cl^-]}{[Br^-]} = \frac{1}{1} = 1$.At standard conditions,the standard cell potential $E_{cell}^{0}$ is determined by the difference in standard electrode potentials of the two half-cells.However,for this specific concentration cell setup,the standard Gibbs free energy change $\Delta G_{cell}^{0}$ is not necessarily zero,and $E_{cell}^{0}$ depends on the standard reduction potentials of $Ag/AgCl/Cl^-$ and $Ag/AgBr/Br^-$.Given the options,$E_{cell} 
eq 0$ because the standard potentials of the two electrodes are different ($E_{Ag/AgCl/Cl^-}^{0} = 0.222 \ V$ and $E_{Ag/AgBr/Br^-}^{0} = 0.071 \ V$).

For the cell: $Ag | AgCl | KCl(1 \ M) || KBr(1 \ M) | AgBr | Ag$,select the correct option.

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