The emf of the cell Ag | Ag^{+}(0.1 , M) || A | vedclass

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(B) The cell reaction is: $Ag^{+}(1 \, M) + Ag_{(s)} \rightarrow Ag_{(s)} + Ag^{+}(0.1 \, M)$.Here,$n = 1$.Using the Nernst equation: $E_{cell} = E_{c

Vedclass · Accepted Answer

$0.059$

Vedclass · Answer

(B) The cell reaction is: $Ag^{+}(1 \, M) + Ag_{(s)} ightarrow Ag_{(s)} + Ag^{+}(0.1 \, M)$.Here,$n = 1$.Using the Nernst equation: $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}$.Since $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.80 \, V - 0.80 \, V = 0 \, V$.$E_{cell} = 0 - \frac{0.0591}{1} \log \frac{0.1}{1}$.$E_{cell} = -0.0591 	imes \log(10^{-1})$.$E_{cell} = -0.0591 	imes (-1) = 0.0591 \, V \approx 0.059 \, V$.

The $emf$ of the cell $Ag | Ag^{+}(0.1 \, M) || Ag^{+}(1 \, M) | Ag$ at $298 \, K$ is ............ $V$.

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