Question

The overall reaction in the lead storage battery is

Pb(s) + PbO2(s) + 2H1(aq) + 2HSO42(aq) ----> 2PbSO4(s) + 2H2O(l)



Calculate % at 258C for this battery when [H2SO4] 5 4.5 M; that is, [H1] 5 [HSO42] 5 4.5 M. At 258C, %8 5 2.04 V for the lead storage battery.

Answer 1

2.12 V

Step-by-step explanation:

see attached file for the detailed step by step solution of the given problem.

Answer 2

Complete Question

The overall reaction in the lead storage battery is

`Pb(s) + PbO2(s) + 2H(aq)^+ + 2HSO_4^-(aq) ----> 2PbSO4(s) + 2H2O(l)`

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
`E = 2.12V`

Step-by-step explanation:

From the question we are told that

The original voltage of the battery is
`E = 2.04 V`

The concentration of [H2SO4] = 5.0 M

The concentration of [H + ] = [HSO4− ] = 5.0 M

At equilibrium

The reaction quotient is

`Q = (1)/([[HSO_4^-]^2 [H^+] ^2])`

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

`Q = (1)/(5^2 * 5^2)`

`= 1.6*10^(-3)`

So the potential for the battery cell is mathematically evaluated as

`E = E^o - (0.05916)/(2) log Q`

`= 2.04 - [(0.05916)/(2) log (1.6*10^(-3))]`

`E = 2.12V`