Question

A digital watch battery draws 0.40 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 1.80 g of each of these four substances, for how many hours will the watch continue to run?

Answer 1




The half reactions for the given reaction are as follows:

Hg^2+ + 2e^- --> Hg reduction
Zn --> Zn^2+ + 2e^- -->Zn oxidation

An ampere is a unit of charge per time (Coulomb per second). There are 2 electrons passing into conduction during this reaction per atom of Zn, ZnO, Hg and HgO present (in a 1:1:1:1 ratio). Each electron has a charge of -e (-1.6x10^-19C).

So, one must determine the limiting reagent by finding out how many mol of each are in 1.00g of each. Use a periodic table to find the molar masses of each (Zn: 65.41g/mol, ZnO: 81.41g/mol, Hg: 200.59g/mol, HgO: 216.59g/mol) and divide the mass m (1.00g) given by the molar mass M to find the number of moles n:

n = m/M

The answer turns out to be 0.00828mol of HgO as the limiting reagent. 0.00828mol of HgO will produce 2*0.008286mol = 0.01656 mol of electrons (2mol of electrons per 1mol of reaction)

Now, the charge calculation:

0.01656mol*6.022x10^23electrons/mol*|-1... = 997.24C = q

Since I = q/t,

t = q/I

t = 997.24/0.40

t = 2.49x10^6sec * 1hr/3600s = 691.67 hrs

Hope this will you.