Question

The mass of an alkaline manganese dioxide AA battery is 25.0 g. Forty percety of the mass is casing and the nominal voltage is +1.5 V. What is the maximum amount of electric energy that can be obtained from this battery?

Answer 1

Mass of whole battery =25 g

Mass casing = 40 % of 25

Mass of magnesium dioxide in battery = 60 % of 25 g

= 60 ×25 /100

= 15 g

Molar mass of mnO2 = 87 g /m

Number of moles of MnO2 = 15 /87 =0.1724 mol

The reduction reaction in battery :

Mn2+ + 2e ---> Mn

From this 1 mol Mn2+ ----> 2 mol electrons

Number of moles of electron inthis case(n) =2×0.1724

= 0.3448 mol

emf of cell (E)=1.5v

1 F = 96500 C

Maximum amount of electric energy obtained from a cell is given by, - change in gibbs energy = nF E

=0.3448× 96500× 1.5

=49913.79 J

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