A current of 3.7 A is passed for 6hrs. Between Ni electrodes in 0.5 L of 2 M solution of `Ni(NO_(3))_(2)`. What will be the molarity of solutionn at the end of electrolysis?

Answer

`1.172M`

Explanation

Given , `I=3.7A`
`t=6xx60xx60s`
`Q=I.t`
`=3.7xx6xx60xx60C=79920C`
`:' 96500C` of electricity deposits `Ni=1Eq`
`:. 79920C` of electricity will deposit `Ni=(1xx79290)/(96500)`
`=0.828 Eq`
Number of moles of `Ni=(Eq of Ni)/(Valency)=(0.828)/(2)=0.414mol`
`[:' ln Ni(NO_(3))_(2),` valency of `Ni=2]`
Volume `(V)=0.5l`
Molarity `(M)=2M`
`2=(Number of mol es)/(0.5)`
Number of moles `=1.0 mol`
Thus, nickel left in` 0.5L` of solution
`=` Number of moles of `Ni` in original solution `-` Number of moles of `Ni` deposited
`=1.0-0.414`
Thus, molarity `(M)=(0.586)/(0.5)=1.172M`