For the cell at 298 K:Ag_{(s)} | AgBr_{(s)} | vedclass

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Question

(B) The cell reaction is: $AgBr_{(s)} + I^-_{(aq)} \rightleftharpoons AgI_{(s)} + Br^-_{(aq)}$The equilibrium constant $K_{eq}$ is given by: $K_{eq} =

Vedclass · Accepted Answer

$E_{cell} = 0.018 \ V$

Vedclass · Answer

(B) The cell reaction is: $AgBr_{(s)} + I^-_{(aq)} ightleftharpoons AgI_{(s)} + Br^-_{(aq)}$The equilibrium constant $K_{eq}$ is given by: $K_{eq} = \frac{K_{sp}(AgBr)}{K_{sp}(AgI)} = \frac{4 	imes 10^{-13}}{8 	imes 10^{-17}} = 0.5 	imes 10^4 = 5000$.$E^o_{cell} = \frac{0.06}{1} \log K_{eq} = 0.06 	imes \log(5000) = 0.06 	imes (3 + \log 5) = 0.06 	imes (3 + 0.7) = 0.06 	imes 3.7 = 0.222 \ V$.Using the Nernst equation: $E_{cell} = E^o_{cell} - 0.06 \log \frac{[Br^-]}{[I^-]} = 0.222 - 0.06 \log \frac{0.01}{0.02} = 0.222 - 0.06 \log(0.5) = 0.222 - 0.06(-0.3) = 0.222 + 0.018 = 0.240 \ V$.$\Delta G^o = -nFE^o_{cell} = -1 	imes 96500 	imes 0.222 \ J$.

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

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For the cell at $298 \ K$:$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$The correct information is:[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]