A digital watch battery draws 0.20 milliamperes of current, which is provided by

Question

A digital watch battery draws 0.20 milliamperes of current, which is provided by a mercury battery whose net reaction is: HgO(s) + Zn(s) ZnO(s) + Hg(l) If a partially used battery contains 1.30 g of each of these four substances, for how many hours will the watch continue to run?

Determine the limiting reagent in terms of moles. Then multiply by the stoichiometric ratio to determine the number of moles of electrons produced. Then convert moles of electons into Coulombs (Amps*seconds). Then you just need to divide by amps used to find the total number of seconds, and convert to hours.

Explanation / Answer

For the same mass i.e. 1.3 g, HgO contains the fewest number of moles (1.3/216.59 = 6 x10-3 moles) and is hence the limiting reagent.

Now looking at the reaction at cathode Hg2+ + 2e Hg(0)

So Number of moles of electrons transferred = 2 (6 x10-3)

Charge on 1 mole of electrons = 1 faraday = 9.648 70 x 104 coulombs

Charge on 2 (6 x10-3) moles of electrons = 1157.84 coulombs

Time for which this cell can work = Charge / Current = 1157.8 / (0.2 x 10-3) = 5789220 s = 1608.11 hours

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