Question

While the discharging of a lead storage battery following reaction take place.

$PbO_2 + Pb + 4H^\oplus + 2SO_4^{2\ominus} \longrightarrow 2PbSO_4 + 2H_2O$ ; E° = 2.01

Calculate the energy (in kJ) obtained from a lead storage battery in which 0.014 mol of lead is consumed.

Assume a constant concentration of 10.M $H_2SO_4$ (log2 = 0.3)

Answer 1

6 kJ

Answer 2

Step 1: Determine the number of electrons (n) transferred in the reaction.

The given reaction is:

$PbO_2 + Pb + 4H^\oplus + 2SO_4^{2\ominus} \longrightarrow 2PbSO_4 + 2H_2O$

In this reaction, Lead (Pb) is oxidized from oxidation state 0 to +2 in $PbSO_4$.

$Pb \longrightarrow Pb^{2+} + 2e^-$

Lead dioxide ($PbO_2$) is reduced from oxidation state +4 to +2 in $PbSO_4$.

$PbO_2 + 4H^\oplus + 2e^- \longrightarrow Pb^{2+} + 2H_2O$

The number of electrons transferred ($n$) is 2.

Step 2: Calculate the concentrations of $H^\oplus$ and $SO_4^{2\ominus}$ ions.

Given that the concentration of $H_2SO_4$ is $10 \, M$. Sulfuric acid dissociates as:

$H_2SO_4 \longrightarrow 2H^\oplus + SO_4^{2\ominus}$

Therefore, $[H^\oplus] = 2 \times 10 \, M = 20 \, M$

And $[SO_4^{2\ominus}] = 1 \times 10 \, M = 10 \, M$

Step 3: Calculate the cell potential (E) using the Nernst equation.

The Nernst equation is:

$E = E^\circ - \frac{0.059}{n} \log Q$

For the given reaction, the reaction quotient $Q$ is:

$Q = \frac{[PbSO_4]^2 [H_2O]^2}{[PbO_2] [Pb] [H^\oplus]^4 [SO_4^{2\ominus}]^2}$

Since $Pb$, $PbO_2$, and $PbSO_4$ are solids, their activities are taken as 1. The activity of water (solvent/product) is also taken as 1.

So, $Q = \frac{1}{[H^\oplus]^4 [SO_4^{2\ominus}]^2}$

Substitute the values: $E^\circ = 2.01 \, V$, $n = 2$, $[H^\oplus] = 20 \, M$, $[SO_4^{2\ominus}] = 10 \, M$.

$E = 2.01 - \frac{0.059}{2} \log \frac{1}{[20]^4 [10]^2}$

$E = 2.01 - \frac{0.059}{2} \log (20^{-4} \times 10^{-2})$

$E = 2.01 - \frac{0.059}{2} \log ((2 \times 10)^{-4} \times 10^{-2})$

$E = 2.01 - \frac{0.059}{2} \log (2^{-4} \times 10^{-4} \times 10^{-2})$

$E = 2.01 - \frac{0.059}{2} \log (2^{-4} \times 10^{-6})$

$E = 2.01 - \frac{0.059}{2} (-4 \log 2 - 6 \log 10)$

Given $\log 2 = 0.3$ and $\log 10 = 1$.

$E = 2.01 - \frac{0.059}{2} (-4 \times 0.3 - 6 \times 1)$

$E = 2.01 - \frac{0.059}{2} (-1.2 - 6)$

$E = 2.01 - \frac{0.059}{2} (-7.2)$

$E = 2.01 + 0.059 \times 3.6$

$E = 2.01 + 0.2124$

$E = 2.2224 \, V \approx 2.22 \, V$

Step 4: Calculate the energy obtained.

The electrical energy obtained from a galvanic cell is given by:

Energy ($W_{elec}$) = $n_{mol}FE$

Where $n_{mol}$ is the total moles of electrons transferred, $F$ is Faraday's constant ($96500 \, C/mol$), and $E$ is the cell potential.

Given that $0.014 \, mol$ of lead (Pb) is consumed. From the stoichiometry of the reaction, 1 mole of Pb corresponds to the transfer of 2 moles of electrons.

So, the total moles of electrons transferred ($n_{mol}$) = $0.014 \, mol \, Pb \times \frac{2 \, mol \, e^-}{1 \, mol \, Pb} = 0.028 \, mol \, e^-$

Energy = $0.028 \, mol \times 96500 \, C/mol \times 2.22 \, V$

Energy = $2702 \times 2.22 \, J$

Energy = $5998.44 \, J$

Converting to kilojoules:

Energy $\approx 6000 \, J = 6 \, kJ$

Explanation of the solution:

1. Identify n: From the balanced reaction, 2 electrons are transferred (Pb goes from 0 to +2, $PbO_2$ from +4 to +2).

2. Calculate ion concentrations: $10 \, M \, H_2SO_4$ yields $20 \, M \, H^\oplus$ and $10 \, M \, SO_4^{2\ominus}$.

3. Apply Nernst Equation: $E = E^\circ - \frac{0.059}{n} \log Q$.

   $Q = \frac{1}{[H^\oplus]^4 [SO_4^{2\ominus}]^2} = \frac{1}{[20]^4 [10]^2}$.

   Substitute values: $E = 2.01 - \frac{0.059}{2} \log (20^{-4} \times 10^{-2}) = 2.01 - \frac{0.059}{2} (-4 \log 2 - 6 \log 10)$.

   Using $\log 2 = 0.3$, $E = 2.01 - \frac{0.059}{2} (-4 \times 0.3 - 6) = 2.01 - \frac{0.059}{2} (-7.2) = 2.01 + 0.2124 = 2.22 \, V$.

4. Calculate Energy: Energy = $n_{mol}FE$.

   For $0.014 \, mol$ of Pb, $n_{mol} = 0.014 \times 2 = 0.028 \, mol$ electrons.

   Energy = $0.028 \times 96500 \times 2.22 = 5998.44 \, J \approx 6 \, kJ$.