Question

The batteries used in automobiles are rechargeable and consist of Pb and PbO2 anodes and
cathodes respectively immersed in 6 M H2SO4 solution. Standard reduction potentials at 25°C
are as follow: E°( PbSO4/Pb) = - 0.35 V and E°( PbO2/PbSO4) = 1.69 V
a) Write the anode, cathode and total reactions when using the battery (i.e., when it is the
galvanic battery it generates electricity and discharge).
b) Find battery potential and standard free energy.
c) If temperature change of battery is given by the following equation:
E(volt) = -0.096 + 1.9 x 10-3
T -3.04 x 10-6 T2
Calculate the equilibrium constant and the thermodynamic functions of the battery (H °,
S °) at 298 K.

Answer 1

Answer:

Explanation:

(a) Write the anode, cathode and total reactions when using the battery (i.e., when it is the  galvanic battery it generates electricity and discharge).

The potential of a cell is given by  

Ecell = Ecathode-Eanode

At Anode,the reaction which occurs is Oxidation.

At cathode,the reaction which occurs is Reduction.

AT ANODE:

Pb + (SO4)2- = PbSO4 +2e-

AT CATHODE:

PbO2 + 4H + + (SO4)2- +2e- = PbSO4 +2H2O

TOTAL REACTION:

Pb +PbO2+ 2(SO4)2- +4H+ =2PbSO4 +2H2O

(b) Find battery potential and standard free energy.

The standard reduction potentials given in the question are as follows:

E(PbSO4/Pb)=-0.35 V

E(PbO2/PbSO4)=1.69 V

The battery potential can be calculated using the given formula which is:

Ecell=Ecathode –Eanode

Ecell=1.69-(-0.35)

Ecell=2.04V

Standard free energy  is the change in the free energy that follows the formation of a mole of a substance from its constituent elements in their standard state.

Standard free energy is given as:

G=-n F Ecell

Where n=2

F=96500C/mol

G=-2*96500*2.04

G=-393.720kJ/mol

(C)  If temperature change of battery is given by the following equation:

E(volt) = -0.096 + 1.9 x 10-3

T -3.04 x 10-6 T2

Calculate the equilibrium constant and the thermodynamic functions of the battery (H °, S °) at 298 K.

Evolt=-0.96 +1.9*10-3T -3.04*10-4 T2

Differentiate E with respect to the time ,

dE/dT =0+1.9*10-3 -6.08*10-6 T

           =1.9*10-3 -6.08*10-6 T

The formula to find the equilibrium constant is:

Log K=nFEcell/2.303RT

Where the values are given as

R=8.14 J/K mol

n=2

F=96500 C/mol

T=25oC

logK =2*96500*2.04/2303*8.314*(273+25)K

        =69.0029

K=antilog(69.0029)

   =1.0067 *(10)69