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(B) $1$. The cell reaction is: $Zn(s) + 2Ag^+(aq) 	o Zn^{2+}(aq) + 2Ag(s)$.$2$. Calculate the standard cell potential: $E^\circ_{cell} = E^\circ_{cat

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$\frac{4}{5.9}$

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(B) $1$. The cell reaction is: $Zn(s) + 2Ag^+(aq) 	o Zn^{2+}(aq) + 2Ag(s)$.$2$. Calculate the standard cell potential: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 	ext{ V} - (-0.76 	ext{ V}) = 1.56 	ext{ V}$.$3$. Apply the Nernst equation at $298 	ext{ K}$: $E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$.$4$. Here, $n = 2$, $[Zn^{2+}] = 1 	ext{ M}$, and $[Ag^+] = x$. Substituting the values: $1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}$.$5$. Simplify the equation: $1.60 = 1.56 - 0.0295 	imes (-2 \log x) = 1.56 + 0.059 \log x$.$6$. Solve for $\log x$: $0.04 = 0.059 \log x$, which gives $\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$.

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