How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI solution. (Cu = 63.5,1F=96500C) Cu2+ + 2e- Cu

It take 0.54 hours to deposit 6.36g of copper

Faraday's Law I

"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis

W = e.i.t / 96500

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

Explanation:

The mole ratio in a balanced equation tells us the number of moles of each reactant needed and how many moles of product will be made.

We are asked to find the mole ratio between A and D for this reaction.

The mole ratio is found by comparing the coefficients, which are the numbers  in front of the formulas.

A has a coefficient of 4 and D has a coefficient of 3.

Therefore, the ratio between A and D is: