Two electrolytic cells containing silver nitrate solution and dilute sulphuric acid solution were connected in series. A steady current of 2.5 amp was
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 ?
1 Answers
(i) Calculate of the amount of electricity consumed
The equation for the reduction of `Ag^(+)` ions is
`{:(Ag^(+)(aq)+e^(-)rarrAg(s)),((107.8g)" "(1F)):}`
For the reduction of 107.8 g `Ag^(+)` ions charge passed =1F
For the reduction of 1.078 g of `Ag^(+)` ions charged passed `=((1 F))/((107.8g))xx(1.078g)=0.01 F`
`=0.01xx96500 C=965 C`
(ii) Calculate of the weight of oxygen gas liberated
Oxygen gas will be liberated in both the cells according to the reaction :
`{:(2H_(2)O(l)rarr4H^(+)(aq)+O_(2)(g)+4e^(-)),(" "(1 mol)" "(4 F)):}`
In the cell containing silver nitrate solution,
On passing 4F of charge, oxygen liberated =32 g
On passing 965 C of charge, oxygen liberated `=(32g)/((4xx96500C))xx(965C)=0.08 g`
The same amount of oxygen gets liberated in the other cell.
`:.` Total amount of oxygen liberated =0.08+0.08=0.16 g