For the cell,Zn_{(s)} | Zn^{2+} (1 M) || Ag^ | vedclass

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(C) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} -

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increases by $0.0296 \ V$

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(C) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} ightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$.Here,$n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.Initially,$Q_1 = \frac{1}{(1)^2} = 1$,so $E_1 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(1) = E^{\circ}_{cell}$.After the concentration change,$Q_2 = \frac{0.1}{(1)^2} = 0.1$.$E_2 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(0.1) = E^{\circ}_{cell} - 0.0296 	imes (-1) = E^{\circ}_{cell} + 0.0296 \ V$.The change in $EMF$ is $E_2 - E_1 = +0.0296 \ V$.Thus,the $EMF$ increases by $0.0296 \ V$.

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

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