What current is required to produce 91.6 g of chromium metal from chromiumVI oxide in 12.4 hours?

Answer 1

Current is defined as the quantity of charge passing a certain point each second given by:

I= Q/t ,where t is time (in seconds) and Q is the charge (in Coulombs).

91.06g of chromium metal (Cr(s)) translates to

n=m/M=91.6g/52.00g/mol= 1.762 mol

Next we must figure out how many electrons are required per reduction reaction.

Cr4+ + 4e- ----> Cr(s), therefore we need 4 electrons per chromium atom to convert it from the oxide to the metal.

#of e- = 4(1.762 mol) = 7.046 mol e-

Lastly we need to convert this number of electrons into a charge (in couloubs).

The charge of one electron is e = 1.6E-19 C, so the total charge is:

Q=(7.046 mol e-)(1.6E-19 C)(6.022E23 e-/mol)= 678900 C

t = (12.4 h)(60min/h)(60s/min)= 44640 s

I= 678900 C/44640 s = 15.2 A

Therefore the current required to produced 91.6 g of chromium metal in 12.4 hours is 15.2 A.