What current (in Ampères) must be provided to deposit the metal at a rate of 6.655 mol/hr if n = 2?HELP ME PLZ

Question

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 6.655 mol/hr if n = 2?

CAN ANYONE HELP ME PLZ!!!!!!!

Answer 1

2 moles of electrons = 1 mole of metal

so you have 6.655*2 = 13.31 moles electrons per hour

Each mole of electrons contains 96490 Coulumbs of charge:

13.31 mol/hr * 96490 C/mol= 1284281.9 Coulumbs / hr

But current is in amperes, which are C/second, so divide the above number by 3600 (seconds/hr) to get the answer:

356.7Amps