CHM 311 practice problems Ch 14-06 An automotive lead-acid battery consists of 6

Question

CHM 311 practice problems Ch 14-06 An automotive lead-acid battery consists of 6 cells that work by the following redox reaction At full charge the sulfuric acid electrolyte is about 4 M in strength. a) Calculate E and E for the cell. b) A fully charged "12-V" battery actually has a somewhat higher voltage. As the battery discharges, sulfuric acid is consumed and the voltage drops, If the battery is discharged to a voltage of less than about 10 V it will likely be impossible to recharge it. What is the molarity ofsul furic acid when battery voltage-10 V (cell voltage = 1.67 V)

Explanation / Answer

a)

E°cell = Ecathode - Eanode

PbO2(s) + 4 H+ + 2 e Pb2+ + 2 H2O +1.460 (cathode)

PbSO4(s) + 2 e Pb(s) + SO42 0.3588 (anode, must be inverted)

Ecell = 1.46 - - 0.3588

E°cell = 1.8188 V

now,

Ecell:

When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.

The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants

The Nernst Equation:

Ecell = E0cell - (RT/nF) x lnQ

In which:

Ecell = non-standard value

E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where

Q = [C]^c * [D]^d / [A]^a*[B]^b

pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)

Q = P-A^a / (P-B)^b

substitute in Nernst Equation:

Ecell = E° - (RT/nF) x lnQ

Q = 1/([H+]^4 * [SO4-2]^2)

4M of H2SO4 = 8M of H+, 4M of SO4-2

Q = 1/((8^4)*(4)^2) = 0.00001525

Ecell = E° - (RT/nF) x lnQ

Ecell = 1.8188 -8.314*298/(4*96500)* ln (0.00001525)

Ecell = 1.89

b)

Find Ecell given, 10 V total

Ecell = E° - (RT/nF) x lnQ

Let "M" be H2SO4, concentration, so H+ = 2M, SO4-2 = M

1.67 = 1.89- (8.314*298/(4*96500) * ln ( 1/(2M)^4 * (M)^2))

1.67 = 1.89- (8.314*298/(4*96500) * ln ( 1/16 / ( M^6))

(1.67 -1.89 ) /((8.314*298/(4*96500)) =  ln ( 1/16 / ( M^6))

exp(-34.275) = 1/16 / ( M^6))

16*(1.30*10^-15) = 1/M^6

M^6 = 1/(16*(1.30*10^-15) ) = 4.8076*10^13

M = (4.8076*10^13)^(1/6) = 190.6 M

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