Calculate Concentration of Silver Ions from Cell Potential

To find the concentration of $\text{Ag}^+$ ions in terms of $\log x$ (where $x = [\text{Ag}^+]$), we use the Nernst equation for the given voltaic cell.

1. Identify the Anode and Cathode Reactions

Since $E^\ominus_{\text{Ag}^+/\text{Ag}} > E^\ominus_{\text{Zn}^{2+}/\text{Zn}}$, silver acts as the cathode (reduction) and zinc acts as the anode (oxidation).

• Anode (Oxidation): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$

• Cathode (Reduction): $2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s)$

• Overall Cell Reaction: $\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)$

Here, the number of electrons involved in the balanced redox reaction is $n = 2$.

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2. Calculate Standard Cell EMF ($E^\ominus_{\text{cell}}$)

$E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ $E^\ominus_{\text{cell}} = E^\ominus_{\text{Ag}^+/\text{Ag}} - E^\ominus_{\text{Zn}^{2+}/\text{Zn}}$ $E^\ominus_{\text{cell}} = 0.80\text{ V} - (-0.76\text{ V}) = 1.56\text{ V}$

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3. Apply the Nernst Equation

The Nernst equation at $298\text{ K}$ is given by: $E_{\text{cell}} = E^\ominus_{\text{cell}} - \frac{0.059}{n} \log Q$

The reaction quotient $Q$ for the cell reaction is: $Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} = \frac{1}{x^2}$

Substituting the given values into the Nernst equation: $1.60 = 1.56 - \frac{0.059}{2} \log \left(\frac{1}{x^2}\right)$

Using the property of logarithms $\log(x^{-2}) = -2\log x$: $1.60 = 1.56 - \frac{0.059}{2} (-2 \log x)$ $1.60 = 1.56 + 0.059 \log x$

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4. Solve for $\log x$

$1.60 - 1.56 = 0.059 \log x$ $0.04 = 0.059 \log x$ $\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$

Thus, the correct option is B.