One half cell in a voltaic cell is constructed by dipping silver rod in $AgNO_3$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $ZnSO_4$.
A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $Ag^+$ ions used in terms of $\log x$ ($x = [Ag^+]$)?
$E^\ominus_{Zn^{2+}/Zn} = -0.76 V$, $E^\ominus_{Ag^+/Ag} = +0.80 V$, $\frac{2.303 RT}{F} = 0.059 V$
Step 1: Understanding the Concept:
We use the Nernst equation to find the cell potential. First, we identify the cathode and anode based on standard reduction potentials, then determine the overall cell reaction and standard EMF. : Key Formula or Approach:
Standard Cell EMF: $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$.
Nernst Equation: $E_{\text{cell}} = E^\ominus_{\text{cell}} - \frac{0.059}{n} \log Q$. Step 2: Detailed Explanation: 1. Identifying Electrodes:
Since $E^\ominus_{Ag^+/Ag} (0.80 V)>E^\ominus_{Zn^{2+}/Zn} (-0.76 V)$, Silver acts as the cathode and Zinc acts as the anode.
Anode (Oxidation): $\text{Zn} \to \text{Zn}^{2+} + 2e^-$
Cathode (Reduction): $2\text{Ag}^+ + 2e^- \to 2\text{Ag}$
Overall reaction: $\text{Zn} + 2\text{Ag}^+ \to \text{Zn}^{2+} + 2\text{Ag}$.
Here, $n = 2$.
2. Calculating Standard EMF:
$E^\ominus_{\text{cell}} = 0.80 V - (-0.76 V) = 1.56 V$.
3. Applying Nernst Equation:
$E_{\text{cell}} = E^\ominus_{\text{cell}} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$
$1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}$
$1.60 - 1.56 = -\frac{0.059}{2} \cdot (-2 \log x)$
$0.04 = 0.059 \log x$
$\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$. Step 3: Final Answer:
The value of $\log [Ag^+]$ is $\frac{4}{5.9}$.
