A digital watch battery draws 0.15 milliamperes of current, which is provided by a mercury battery whose net reaction is: HgO(s) + Zn(s) → ZnO(s) + Hg(l)

Question

If a partially used battery contains 1.70 g of each of these four substances, for how many hours will the watch continue to run?

Answer 1

HgO(s) + Zn(s) → ZnO(s) + Hg(l)

As Mass of HgO = 1.70 g

Molar Mass of HgO = 216.59 g/mol

So , Moles of HgO = 1.70 / 216.59 = 0.007849 mol

and, Mass of Zn = 1.70 g

Molar Mass of Zn = 65.38 g/mol

So , Moles of Zn = 1.70 / 65.38 = 0.02600 mol

As 1 mole of HgO combines with 1 mole of Zn.

So, 0.007849 moles of HgO reacts with 0.007849 moles of Zn. But there are 0.02600 moles of Zn.

So, HgO is the limiting reactant.

Now, Half - the cell reaction :

Hg2+ + 2e- ----> Hg

So, no. of electrons transferred = Z = 2

Faraday's constant = F = 96500 C / mol

moles of Hg2+ = n = 0.007849 mol

Current = I = 0.15 mA = 0.15 × 10^-3 A

So, Time in which the watch continues to run = t

= n Z F / I

= 0.007849 × 2 × 96500 / (0.15 × 10^-3 )

= 1.0099046×10^7 s

= 1.0099046×10^7 / 3600 hr

= 2805.29055 hr