# Answer to Question #42832 in Physical Chemistry for anuragasaksena

Question #42832

The cathodic reaction of a dry cell is represented by 2MnO2(s) + Zn2+ + 2e- → ZnMn2O4(s) If, there are 8 g of MnO2 in the cathodic compartment then the time for which the dry cell will continue to give a current of 2 milliampere is:

Expert's answer

According to the Faraday's law:

m = (It/F)*M/z

where m is the mass of substance, I is the current strength, t is the time of experiment, F is the Faraday's constant, M is the molar mass of substance, z is the factor of equivalence.

The time of the electrolysis can be expressed as follows:

t = mzF/IM

The factor of equivalence can be found dividing the number of electrons by the number of reacted particles. In case of MnO

z = 2/2 = 1

The molar mass of MnO

M(MnO

Now we have enough data to calculate the time:

t = 8*1*96485/(0.002*87) = 4.436*10

These calculations were made in assumption that the current strength will be constant until the full amount of MnO

m = (It/F)*M/z

where m is the mass of substance, I is the current strength, t is the time of experiment, F is the Faraday's constant, M is the molar mass of substance, z is the factor of equivalence.

The time of the electrolysis can be expressed as follows:

t = mzF/IM

The factor of equivalence can be found dividing the number of electrons by the number of reacted particles. In case of MnO

_{2}the factor is:z = 2/2 = 1

The molar mass of MnO

_{2}is:M(MnO

_{2}) = 55 + 16*2 = 87 g/molNow we have enough data to calculate the time:

t = 8*1*96485/(0.002*87) = 4.436*10

^{6}s = 1232 hours = 51.3 daysThese calculations were made in assumption that the current strength will be constant until the full amount of MnO

_{2}is consumed. However in real life the current strength will gradually drop with the consumption of reactants, so the real time of use will be lower.
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