(a) Two electrolytic cells containing silver nitrate solution and dilute sulphuric acid solution were connected in series. A steady current of 2.5 amp was passed through them till 1.078 g of silver was deposited. [Ag = 107.8 g `mol^(-1)`, 1 F = 96,500 C]
(i) How much electricity was consumed ?
(ii) What was the weight of oxygen gas liberated ?
(b) Give reason :
(i) Rusting of iron pipe can be prevented by joining it with a piece of magnesium.
(ii) Conductivity of an electrolyte solution decreases with the decrease in concentration.

Explanation

(a) (i) `Ag^(+) + e^(-) to Ag(s)`
Thus, 107.8 g of silver is deposited by = 96500 coulombs
1.078 g of silver is deposited by `=96500/107.8 xx 1.078 = 965` coulombs
(ii) Oxygen is liberated at the anode of the first cell as well as the second cell. Weight of oxygen liberated
`2H_(2)O (l) to O_(2)(g) + 4H^(+) (aq) + 4e^(-)`
4 x 96500 coulombs (4 Faraday) electricity is consumed to liberate 1 mole (32 g) of oxygen. Oxygen liberated with 965 coulombs of electricity
`=32/(4 xx 96500) xx 965 = 0.08 g`
Weight of oxygen liberated at the anode of the two cells = 2 x 0.08 = 0.16 g
(b) (i) Rusting of iron can be prevented by joining iron pipe with a piece of magnesium, llus is called sacrificial method in which magnesium itself, being more reactive, corrodes itself and saves iron pipes.
(ii) Conductivity of an electrolyte solution decreases with decrease of concentration because the number of ions per unit volume decreases with dilution i.e., decrease in concentration.