Questions: What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+]=0.75 M, [H+]=0.063 M and PH2=1.0 atm? Round your answer to 2 significant digits.

What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt / H+ / H2 half-cell if [Pb2+]=0.75 M, [H+]=0.063 M and PH2=1.0 atm? Round your answer to 2 significant digits.

Transcript text: What is the emf of a cell consisting of a $\mathrm{Pb}^{2+} / \mathrm{Pb}$ half-cell and a $\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}$ half-cell if $\left[\mathrm{Pb}^{2+}\right]=0.75 \mathrm{M}$, $\left[\mathrm{H}^{+}\right]=0.063 \mathrm{M}$ and $\mathrm{P}_{\mathrm{H}_{2}}=1.0 \mathrm{~atm}$ ? Round your answer to 2 significant digits.

Solution

Solution Steps

Step 1: Identify the Half-Reactions and Standard Reduction Potentials

The cell consists of two half-cells:

$\mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb}$ with a standard reduction potential $E^\circ = -0.13 \, \text{V}$.
$\mathrm{2H}^+ + 2e^- \rightarrow \mathrm{H}_2$ with a standard reduction potential $E^\circ = 0.00 \, \text{V}$.

Step 2: Determine the Cell Reaction and Standard Cell Potential

The overall cell reaction is: \[ \mathrm{Pb}^{2+} + \mathrm{H}_2 \rightarrow \mathrm{Pb} + 2\mathrm{H}^+ \]

The standard cell potential $E^\circ_{\text{cell}}$ is calculated as: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 \, \text{V} - (-0.13 \, \text{V}) = 0.13 \, \text{V} \]

Step 3: Apply the Nernst Equation

The Nernst equation is used to calculate the cell potential under non-standard conditions: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]

Where:

$R = 8.314 \, \text{J/mol K}$
$T = 298 \, \text{K}$
$n = 2$ (number of electrons transferred)
$F = 96485 \, \text{C/mol}$
$Q = \frac{[\mathrm{H}^+]^2}{[\mathrm{Pb}^{2+}] \cdot P_{\mathrm{H}_2}}$

Substitute the given concentrations and pressure: \[ Q = \frac{(0.063)^2}{0.75 \cdot 1.0} = \frac{0.003969}{0.75} = 0.005292 \]

Substitute into the Nernst equation: \[ E_{\text{cell}} = 0.13 \, \text{V} - \frac{(8.314)(298)}{(2)(96485)} \ln(0.005292) \]

Calculate the term: \[ \frac{(8.314)(298)}{(2)(96485)} \approx 0.01285 \, \text{V} \]

Calculate $\ln(0.005292)$: \[ \ln(0.005292) \approx -5.241 \]

Substitute back: \[ E_{\text{cell}} = 0.13 \, \text{V} - (0.01285 \, \text{V})(-5.241) \approx 0.13 \, \text{V} + 0.0673 \, \text{V} = 0.1973 \, \text{V} \]