Question

A solution of $Ni{{\left( N{{O}_{3}} \right)}_{2}}$ is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many moles Ni will be deposited at the cathode?
  (A) 0.20
  (B) 0.05
  (C) 0.10
  (D) 0.15

Answer 1

Hint: When we pass one Faraday of charge in electrolysis, we are passing one mole of electrons and also, by passing one Faraday of charge, one gram equivalent weight of the substance will be deposited or liberated on the electrode. From this relation we would be able to calculate the number of moles Ni getting deposited at the cathode.

Complete step by step solution:
- Let’s start with the Faradays second law of electrolysis. According to Faraday's second law of electrolysis, the mass of a substance liberated or deposited at any electrode on passing a certain amount of charge will be directly proportional to its chemical equivalent weight.
- The gram-atomic weight can be defined as the mass of one mole of an element equal in grams to the atomic weight and the equivalent weight is obtained by dividing the molar mass by the valency. When one faraday electricity is passed, it will deposit one gram equivalent of the metal.
- The electrolysis of solution of $Ni{{\left( N{{O}_{3}} \right)}_{2}}$ can be represented as follows
\[Ni{{(N{{O}_{3}})}_{2}}\to N{{i}^{2+}}+2NO_{3}^{-}\]
Here the nickel gets oxidized and the corresponding reaction can be written as
\[N{{i}^{2+}}+2{{e}^{-}}\to Ni\]
- As we mentioned above, 2 equivalent of $N{{i}^{2+}}$ will be discharged from 2 Faradays of electricity. Therefore, from 0.1F Faraday electricity 0.1 equivalent of $N{{i}^{2+}}$ will be discharged. This relation can be given as follows,
\[Gram\text{ }equivalent=~n\text{ }Factor\times No.\text{ }of\text{ }moles~\]
Where n Factor is the electrons that get exchanged. We have the value of gram equivalent as 0.1 and the number of electrons involved is two. Substitute these values in the above equation,
\[0.1=~~No.\text{ }of\text{ }moles\text{ }of\text{ }Ni\times 2\]
\[~~No.\text{ }of\text{ }moles\text{ }of\text{ }Ni=\dfrac{0.1}{2}=0.05mol\]
Hence the number of moles Ni getting deposited at the cathode is 0.05 moles.

Therefore the answer is option (B) 0.05.

Note: Keep in mind that the anode is the electrode at which the oxidation reaction occurs while cathode is the electrode where the reduction reaction takes place. Also, do not correlate the positive or negative sign of the electrode with the nature of that electrode. Also it should be noted that the unit of electric charge is Coulomb.