What happens to the emf of the cell Zn_{(s)} | vedclass

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(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} \rightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \fra

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decrease by $0.0592 \ V$

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(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} ightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q$.Here,$n = 2$ and $Q = \frac{[Zn^{+2}]}{[Ag^{+1}]^2}$.Initially,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 1 \ M$,so $Q_1 = \frac{1}{1^2} = 1$. Thus,$E_1 = E^0_{cell} - \frac{0.0592}{2} \log(1) = E^0_{cell}$.Finally,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 0.1 \ M$,so $Q_2 = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.$E_2 = E^0_{cell} - \frac{0.0592}{2} \log(100) = E^0_{cell} - 0.0296 	imes 2 = E^0_{cell} - 0.0592 \ V$.The change in $emf$ is $E_2 - E_1 = -0.0592 \ V$,which means the $emf$ decreases by $0.0592 \ V$.

What happens to the $emf$ of the cell $Zn_{(s)} | Zn^{+2} (1 \ M) || Ag^{+1} (1 \ M) | Ag_{(s)}$ if the concentration of $Ag^{+1}$ decreases to $0.1 \ M$?

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