Question

A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contains $\text{AgN}{\text{O}}_{\text{3}}$ solution and second contains $\text{CuC}{\text{l}}_2$ solution. During this time 4 grams of Ag are deposited in the first cell.
(a) How many grams of Cu are deposited in the second cell?
(b) What is the current flowing in amperes?

Answer 1

1: By Faraday's second law

w $\alpha$ E

$\frac{(w{)}_{Ag}}{(w{)}_{Cu}}$ = $\frac{(E{)}_{Ag}}{(E{)}_{Cu}}$

$\frac{4}{(w{)}_{Cu}}=\frac{107.86}{1}\times \frac{2}{63.546}$

$(w{)}_{Cu}=1.178gm$

2: w = zit

$i=\frac{4\times 1\times 96500}{4\times 3600\times 107.86}$

i = 0.248521 A