chem question?

Question

could anyone tell me how to solve the following problem?

A metal undergoes electrolytic reduction according to
ne- + Mn+ ===> M. What current (in Ampères) must be provided to deposit the metal at a rate of 1.801 mol/hr if n = 1?

Answer 1

From the reaction:
ne- + M(n+) ===> M,
we know that we need to provide n moles of electrons to react with M(n+), in order to get one mole of M.
Recall that the number of particles in one mole is the Avogadro number: 6.022x10^23/mol, and the charge of one electron is: -1.602x10^-19 C. The product of these two number is one Faraday: 96485 C/mol. Also please recall 1A (Ampere) = 1C/sec. Hence the required current can be easily calculated as:
1.801*n*96485 C/3600sec
when n = 1, this number becomes:
(1.801*96485/3600)A = 48.27A (answer)