Question

A cell is constructed as:

$Ag|AgCl_{(s)}|Cl^-(0.1M)||Pb^{2+}_{(sat.soln.ofPbCl_2)}|Pb$

Calculate the EMF of the cell.

• $K_{sp}(AgCl) = 1.6 \times 10^{-10}$
• $K_{sp}(PbCl_2) = 1.7 \times 10^{-5}$
• $E_{Ag^+/Ag}^\circ = +0.80, V$
• $E_{Pb^{2+}/Pb}^\circ = -0.13, V$
• $[Cl^-] = 0.1 M$

Answer 1

-0.46 V

Answer 2

To calculate the EMF of the cell, we need to determine the potential of each half-cell using the Nernst equation and then apply the formula $E_{cell} = E_{cathode} - E_{anode}$.

The given cell notation is: $Ag|AgCl_{(s)}|Cl^-(0.1M)||Pb^{2+}_{(sat.soln.ofPbCl_2)}|Pb$

1. Anode (Left Half-Cell): $Ag|AgCl_{(s)}|Cl^-(0.1M)$

The electrode reaction is $AgCl_{(s)} + e^- \rightleftharpoons Ag_{(s)} + Cl^-$.

First, we need to calculate the standard electrode potential $E^\circ_{AgCl/Ag, Cl^-}$. We know the standard potential for $Ag^+/Ag$ and the $K_{sp}$ for $AgCl$.

The relevant reactions are:

(1) $Ag^+_{(aq)} + e^- \rightleftharpoons Ag_{(s)}$; $E^\circ_1 = E^\circ_{Ag^+/Ag} = +0.80 \, V$

(2) $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$; $K_{sp} = 1.6 \times 10^{-10}$

Adding reactions (1) and (2) gives the desired reaction:

$AgCl_{(s)} + e^- \rightleftharpoons Ag_{(s)} + Cl^-_{(aq)}$

The standard free energy change for the overall reaction is the sum of the standard free energy changes for (1) and (2):

$\Delta G^\circ_{overall} = \Delta G^\circ_1 + \Delta G^\circ_2$

$-nFE^\circ_{overall} = -nFE^\circ_1 - RT \ln K_{sp}$

For this reaction, $n=1$.

$-F E^\circ_{AgCl/Ag, Cl^-} = -F E^\circ_{Ag^+/Ag} - RT \ln K_{sp}$

Dividing by $-F$:

$E^\circ_{AgCl/Ag, Cl^-} = E^\circ_{Ag^+/Ag} + \frac{RT}{F} \ln K_{sp}$

Using $\frac{RT}{F} = 0.0591$ at 298 K and $\ln x = 2.303 \log x$:

$E^\circ_{AgCl/Ag, Cl^-} = E^\circ_{Ag^+/Ag} + 0.0591 \log K_{sp}$

$E^\circ_{AgCl/Ag, Cl^-} = 0.80 + 0.0591 \log (1.6 \times 10^{-10})$

$\log (1.6 \times 10^{-10}) = \log 1.6 - 10 = 0.2041 - 10 = -9.7959$

$E^\circ_{AgCl/Ag, Cl^-} = 0.80 + 0.0591 \times (-9.7959)$

$E^\circ_{AgCl/Ag, Cl^-} = 0.80 - 0.5790 = 0.2210 \, V$

Now, calculate the potential of the anode using the Nernst equation:

$E_{anode} = E^\circ_{AgCl/Ag, Cl^-} - \frac{0.0591}{1} \log \frac{[Ag_{(s)}][Cl^-]}{[AgCl_{(s)}]}$

Since $Ag_{(s)}$ and $AgCl_{(s)}$ are solids, their activities are 1.

$E_{anode} = E^\circ_{AgCl/Ag, Cl^-} - \frac{0.0591}{1} \log [Cl^-]$

Given $[Cl^-] = 0.1 \, M$.

$E_{anode} = 0.2210 - 0.0591 \log (0.1)$

$E_{anode} = 0.2210 - 0.0591 \times (-1)$

$E_{anode} = 0.2210 + 0.0591 = 0.2801 \, V$

2. Cathode (Right Half-Cell): $Pb^{2+}_{(sat.soln.ofPbCl_2)}|Pb$

The electrode reaction is $Pb^{2+}_{(aq)} + 2e^- \rightleftharpoons Pb_{(s)}$.

First, find the concentration of $Pb^{2+}$ in a saturated solution of $PbCl_2$.

PbCl_2_{(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}

Given $K_{sp}(PbCl_2) = 1.7 \times 10^{-5}$.

Let $[Pb^{2+}] = s$. Then $[Cl^-] = 2s$.

$K_{sp} = [Pb^{2+}][Cl^-]^2 = s(2s)^2 = 4s^3$

$s = \left(\frac{K_{sp}}{4}\right)^{1/3} = \left(\frac{1.7 \times 10^{-5}}{4}\right)^{1/3} = (0.425 \times 10^{-5})^{1/3} = (4.25 \times 10^{-6})^{1/3}$

$s = (4.25)^{1/3} \times 10^{-2} \approx 1.6206 \times 10^{-2} \, M$

So, $[Pb^{2+}] = 1.6206 \times 10^{-2} \, M$.

Now, calculate the potential of the cathode using the Nernst equation:

$E_{cathode} = E^\circ_{Pb^{2+}/Pb} - \frac{0.0591}{2} \log \frac{[Pb_{(s)}]}{[Pb^{2+}]}$

Since $Pb_{(s)}$ is a solid, its activity is 1.

$E_{cathode} = E^\circ_{Pb^{2+}/Pb} - \frac{0.0591}{2} \log \frac{1}{[Pb^{2+}]}$

Given $E^\circ_{Pb^{2+}/Pb} = -0.13 \, V$.

$E_{cathode} = -0.13 - \frac{0.0591}{2} \log \frac{1}{1.6206 \times 10^{-2}}$

$E_{cathode} = -0.13 - 0.02955 \log (\frac{100}{1.6206})$

$E_{cathode} = -0.13 - 0.02955 \log (61.705)$

$\log (61.705) \approx 1.7903$

$E_{cathode} = -0.13 - 0.02955 \times 1.7903$

$E_{cathode} = -0.13 - 0.05289 = -0.18289 \, V$

3. Calculate the EMF of the Cell:

$E_{cell} = E_{cathode} - E_{anode}$

$E_{cell} = -0.18289 - 0.2801$

$E_{cell} = -0.46299 \, V$

Rounded to two decimal places, $E_{cell} = -0.46 \, V$.

The negative EMF indicates that the reaction as written (oxidation at Ag electrode and reduction at Pb electrode) is non-spontaneous. The spontaneous reaction would be the reverse.

Explanation of the solution:

1. Calculate Standard Potential for AgCl/Ag, Cl- Electrode:

   The standard potential for the reaction $AgCl_{(s)} + e^- \rightleftharpoons Ag_{(s)} + Cl^-$ is derived from $E^\circ_{Ag^+/Ag}$ and $K_{sp}(AgCl)$ using the relationship:

   $E^\circ_{AgCl/Ag, Cl^-} = E^\circ_{Ag^+/Ag} + 0.0591 \log K_{sp}(AgCl)$

   $E^\circ_{AgCl/Ag, Cl^-} = 0.80 + 0.0591 \log (1.6 \times 10^{-10}) = 0.80 + 0.0591 \times (-9.7959) = 0.2210 \, V$

2. Calculate Potential of Anode (Ag|AgCl, Cl-) Electrode:

   Using the Nernst equation for the reduction reaction $AgCl_{(s)} + e^- \rightleftharpoons Ag_{(s)} + Cl^-$:

   $E_{anode} = E^\circ_{AgCl/Ag, Cl^-} - 0.0591 \log [Cl^-]$

   Given $[Cl^-] = 0.1 \, M$:

   $E_{anode} = 0.2210 - 0.0591 \log (0.1) = 0.2210 - 0.0591 \times (-1) = 0.2210 + 0.0591 = 0.2801 \, V$

3. Calculate Concentration of Pb2+ in Cathode (Pb2+|Pb) Electrode:

   The cathode involves a saturated solution of $PbCl_2$.

   For PbCl_2_{(s)} \rightleftharpoons Pb^{2+} + 2Cl^-, $K_{sp} = [Pb^{2+}][Cl^-]^2$.

   Let $[Pb^{2+}] = s$, then $[Cl^-] = 2s$.

   $K_{sp} = s(2s)^2 = 4s^3$

   $s = \left(\frac{K_{sp}}{4}\right)^{1/3} = \left(\frac{1.7 \times 10^{-5}}{4}\right)^{1/3} = (4.25 \times 10^{-6})^{1/3} \approx 1.6206 \times 10^{-2} \, M$

   So, $[Pb^{2+}] = 1.6206 \times 10^{-2} \, M$.

4. Calculate Potential of Cathode (Pb2+|Pb) Electrode:

   Using the Nernst equation for the reduction reaction $Pb^{2+} + 2e^- \rightleftharpoons Pb_{(s)}$:

   $E_{cathode} = E^\circ_{Pb^{2+}/Pb} - \frac{0.0591}{2} \log \frac{1}{[Pb^{2+}]}$

   Given $E^\circ_{Pb^{2+}/Pb} = -0.13 \, V$:

   $E_{cathode} = -0.13 - \frac{0.0591}{2} \log \frac{1}{1.6206 \times 10^{-2}}$

   $E_{cathode} = -0.13 - 0.02955 \log (61.705) = -0.13 - 0.02955 \times 1.7903 = -0.13 - 0.05289 = -0.18289 \, V$

5. Calculate EMF of the Cell:

   $E_{cell} = E_{cathode} - E_{anode}$

   $E_{cell} = -0.18289 \, V - 0.2801 \, V = -0.46299 \, V$

Rounded to two decimal places, $E_{cell} = -0.46 \, V$.