0.1 F charge is supplied to a solution of $Ni{{(N{{O}_{3}})}_{2}}$. The amount of Ni deposited (in mol) at the cathode will be:
(a)- 0.05
(b)- 1.0
(c)- 0.5
(d)- 0.10

Hint: The amount of the nickel deposited at the cathode can be calculated by dividing the number charge supplied to the solution to the number of electrons involved in the reaction. Nickel in solution has a +2 oxidation state.

Complete step by step answer:
When charge is passed through the solution of $Ni{{(N{{O}_{3}})}_{2}}$, the $Ni{{(N{{O}_{3}})}_{2}}$will split into ions. The ions formed will be nickel ions and nitrate ions. The reaction is given below:
$Ni{{(N{{O}_{3}})}_{2}}\to N{{i}^{2+}}+2NO_{3}^{-}$
So, in this electrolysis, we can see that the number of electrons involved in this reaction is 2. Or we can say that the nickel-metal $(Ni)$ loses 2 electrons and these 2 electrons are gained by 2 molecules of $N{{O}_{3}}$.
At the cathode, oxidation of nickel takes and the reaction is:
$N{{i}^{2+}}+2{{e}^{-}}\to Ni$
And we know that when n electrons are involved in the reaction then the passage of n faraday of current will produce 1 mole of a substance.
So, for the above reaction, we can say that 2 electrons are involved in the reaction then the passage of 2 faradays will produce 1-mole substance.
$2F\to 1\text{ }mole$
When 0.1 F of current is passed through the solution then the amount of nickel produced will be:
$0.1F\to \dfrac{0.1}{2}=0.05\text{ moles}$
So, 0.05 moles of nickel will be produced at the cathode.

Therefore the correct answer is an option (a)- 0.05.


Note: In terms of gram equivalent, it may be remembered that One faraday (i.e., 96500 coulombs) of current will deposit one gram equivalent of the substance. The equivalent weight is equal to the ratio of the atomic weight of the element to the electrons gained or lost by the atom or ion.