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# 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:
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2014-05-26T14:07:05-0400
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&#039;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 MnO2 the factor is:
z = 2/2 = 1
The molar mass of MnO2 is:
M(MnO2) = 55 + 16*2 = 87 g/mol
Now we have enough data to calculate the time:
t = 8*1*96485/(0.002*87) = 4.436*106 s = 1232 hours = 51.3 days
These calculations were made in assumption that the current strength will be constant until the full amount of MnO2 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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