Complete Question
The overall reaction in the lead storage battery is
[tex]Pb(s) + PbO2(s) + 2H(aq)^+ + 2HSO_4^-(aq) ----> 2PbSO4(s) + 2H2O(l)[/tex]
Calculate E at 25°C for this battery when [H2SO4] = 5.0 M; that is, [H + ] = [HSO4− ] = 5.0 M. At 25°C, E° = 2.04 V for the lead storage battery.
Answer:
The voltage of the cell is [tex]E = 2.12V[/tex]
Explanation:
From the question we are told that
The original voltage of the battery is [tex]E = 2.04 V[/tex]
The concentration of [H2SO4] = 5.0 M
The concentration of [H + ] = [HSO4− ] = 5.0 M
At equilibrium
The reaction quotient is
[tex]Q = \frac{1}{[[HSO_4^-]^2 [H^+] ^2]}[/tex]
Pb(s), Pb(s), 2 H2O(l),2 PbSO4(s) are excluded from the reaction above because they are solid and liquid thus there concentration does not change
[tex]Q = \frac{1}{5^2 * 5^2}[/tex]
[tex]= 1.6*10^{-3}[/tex]
So the potential for the battery cell is mathematically evaluated as
[tex]E = E^o - \frac{0.05916}{2} log Q[/tex]
[tex]= 2.04 - [\frac{0.05916}{2} log (1.6*10^{-3})][/tex]
[tex]E = 2.12V[/tex]
