Two electrolytic cells containing AgNo3 solution and dil.H2SO4 solution are connected in series.A steady current of 2.5 amp was passed till 1.078g of silver was deposited.

a)How much electricity was produced?

b)What was the weight of O2 liberated?

a. As silver is deposited on the electrode from AgNO₃ solution, this is  a reduction reaction  

Ag⁺(aq) + e^- $\to$  Ag(s) 

Moles of Ag depositied  =  mass deposited/atomic mass of Ag =  1.078/107.8 or 0.01mol

(atomic mass of Ag is 107.8u)

As, 1 mol of silver ions require 1mol of electrons for reduction.

Moles of electrons required for reduction of 1.078g of silver ions is  0.01mol

Electricity required to deposit of 0.01mol Ag = 0.01F = 0.01 $\times$ 96500 C = 965C

b. In the second cell water is oxidised to oxygen  2H₂O (aq) $\to$    O₂  +  4H⁺  +  4e^-

Therefore, 4 moles of electrons are required to deposit 1mol O₂

0.01mol of electrons will deposit  = $\frac{1\times 0.01}{4}$

                                                 = 0.0025mol of O₂

Mass of oxygen liberated  =  moles $\times$ molar mass = 0.0025 $\times$ 32 = 0.08g