Question

The overall reaction in the lead storage battery is
$\mathbf{Pb}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{PbO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{HS}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{PbSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$
a. Calculate ${\mathbf{E}}_{\mathbf{cell}}$at 25°C for this battery when [H₂SO₄] = 4.5 M; that is, [H⁺] = [HSO₄^-] = 4.5 M. At 25°C, E° = 2.04 V for the lead storage battery.
b. For the cell reaction ${\mathbf{\Delta H}}^{\mathbf{o}}$= -315.9 kJ and${\mathbf{\Delta S}}^{\mathbf{o}}$= 263.5 J/K. Calculate E° at -20.ºC. (See Exercise 45.)
c. Calculate ${\mathbf{E}}_{\mathbf{cell}}$at -20.°C when [H₂SO₄] = 4.5 M.
d. Based on your previous answers, why does it seem that batteries fail more often on cold days than on warm days?

Answer

Short answer

(a) The value will be${\mathrm{E}}_{\mathrm{cell}}=2.12 \mathrm{V}$

(b) The value will be${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=1.98 \mathrm{V}$

(c) The value will be${\mathrm{E}}_{\mathrm{cell}}=2.05 \mathrm{V}$

(d)There is an increase in electrode potential and decrease in cell potential temperature as both are inversely proportional to each other. When cold days the temperature will be low, then the cell potential will increase. Thus, there is more viscous of oil in battery and no current flow. Therefore, batteries will fail on cold days more often than warm days.

Step-by-step solution

Step 1(a): Finding electrode potential of the cell using Nernst equation

On using Nernst equation,

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \left(\frac{1}{{\left[\mathrm{HS}{{\mathrm{O}}_4}^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^2}\right)\\ {\mathrm{E}}_{\mathrm{cell}}& =& 2.04-\frac{0.0591}{\mathrm{n}}\log \left(\frac{1}{{[4.5]}^2{[4.5]}^2}\right)\\ {\mathrm{E}}_{\mathrm{cell}}& =& 2.04+0.077\\ {\mathrm{E}}_{\mathrm{cell}}& =& 2.12 \mathrm{V}\end{array}$

Step 2(b): Finding change in standard Gibbs free energy

On using the formula,

$\begin{array}{rcl}{\mathrm{\Delta G}}^{\mathrm{o}}& =& {\mathrm{\Delta H}}^{\mathrm{o}}-{\mathrm{T\Delta S}}^{\mathrm{o}}\\ {\mathrm{\Delta G}}^{\mathrm{o}}& =& (-315.9\mathrm{kJ})-(253.15 \mathrm{K})(263.5 \mathrm{J}/\mathrm{k})\left(\frac{{10}^{-3}\mathrm{kJ}}{1 \mathrm{J}}\right)\\ {\mathrm{\Delta G}}^{\mathrm{o}}& =& -382.6 \mathrm{kJ}\end{array}$

Then substituting the values to find standard electrode potential,

$\begin{array}{rcl}{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& -\frac{{\mathrm{\Delta G}}^{\mathrm{o}}}{\mathrm{nF}}\\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& \frac{-(-382.6 \mathrm{kJ})}{2(96485 \mathrm{C}/\mathrm{mol})}\left(\frac{{10}^{-3} \mathrm{J}}{1 \mathrm{kJ}}\right)\\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& 1.98 \mathrm{V}\end{array}$

Step 3(c): Calculating electrode potential of the cell at -20°C

On change temperature unit,

$\begin{array}{rcl}\mathrm{T}\left(\mathrm{K}\right)& =& \mathrm{T}\left({}^{\mathrm{o}}\mathrm{C}\right)+273.15\\ \mathrm{T}\left(\mathrm{K}\right)& =& -20+273.15\\ \mathrm{T}\left(\mathrm{K}\right)& =& 253.15 \mathrm{K}\end{array}$

The Nernst equation can be expressed as,

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{\mathrm{RT}}{\mathrm{nF}}\left(\frac{1}{\left[\mathrm{HS}{{\mathrm{O}}_4}^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^2}\right)\\ {\mathrm{E}}_{\mathrm{cell}}& =& 2.04-\frac{(8.314 \mathrm{J}/\mathrm{K} \mathrm{mol})(253.15 \mathrm{K})}{2(96485 \mathrm{C}/\mathrm{mol})}\log \left(\frac{1}{{(4.5)}^2{(4.5)}^2}\right)\\ {\mathrm{E}}_{\mathrm{cell}}& =& 2.05 \mathrm{V}\end{array}$

Step 4(d): Batteries fail more often on cold days than on warm days

The expression of Nernst equation can be as follows,

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{logQ}$

There is an increase in electrode potential and decrease in cell potential temperature as both are inversely proportional to each other. When cold days the temperature will be low, then the cell potential will increase. Thus, there is more viscous of oil in battery and no current flow. Therefore, batteries will fail on cold days more often than warm days.