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Question: E.M.F of \(Ni(s)/N{{i}^{+2}}(aq)\parallel C{{u}^{+2}}/Cu(s)\) cell can be increased by (A) Adding...

E.M.F of Ni(s)/Ni+2(aq)∥Cu+2/Cu(s)Ni(s)/N{{i}^{+2}}(aq)\parallel C{{u}^{+2}}/Cu(s) cell can be increased by
(A) Adding NH3N{{H}_{3}} in the right half-cell
(B) Increasing the concentration of Ni+2N{{i}^{+2}} ions
(C) Adding dimethylglyoxime into the left half-cell
(D) Changing the electrolyte present in the salt bridge

Explanation

Solution

The maximum potential difference between two electrodes of a cell is known as the electromotive force of a cell or EMF of a cell which is used to determine if an electrochemical is a galvanic cell or not. This EMF of a cell is also referred to as the net voltage between the oxidation and reduction half-reactions.

Complete step by step solution:
A potential difference is set up across the surface when a metal electrode is immersed in a solution containing its ions and this potential difference is known as electrode potential.
The cell potential is also referred to as EMF of a cell and can be calculated by considering the values of electrode potential of two half cell reactions. Three methods can be used for this calculation:
(i) The oxidation potential of the anode and reduction potential of the cathode
(ii) The reduction potentials of both electrodes, or
(iii) The oxidation potentials of both electrodes
The cell of the potential of a galvanic cell = potential of the half cell on the right side (cathode) – the potential of the half cell on the left (anode)
Given cell reaction,
Ni(s)/Ni+2(aq)∥Cu+2/Cu(s)Ni(s)/N{{i}^{+2}}(aq)\parallel C{{u}^{+2}}/Cu(s)
Cathode: Cu2++2e−→CuC{{u}^{2+}}+2{{e}^{-}}\to Cu
Anode: Ni→Ni+2+2e−Ni\to N{{i}^{+2}}+2{{e}^{-}}
Net reaction: Cu+2+Ni→Cu+Ni+2C{{u}^{+2}}+Ni\to Cu+N{{i}^{+2}}
Nernst equation for this cell reaction,
Ecell=Ecell0−0.05912log⁡[Ni+2][Cu+2]{{E}_{cell}}=E_{cell}^{0}-\dfrac{0.0591}{2}\log \dfrac{[N{{i}^{+2}}]}{[C{{u}^{+2}}]}
If we add dimethylglyoxime in the left-half cell, then dimethylglyoxime will bind the Ni+2N{{i}^{+2}} which results in the concentration of free Ni+2N{{i}^{+2}} in solution. Hence, EMF of a given cell increases by adding dimethylglyoxime into the left half-cell.

So, the correct answer is option C.

Note: A device that generates electricity from a chemical reaction is an electrochemical cell and also this device is used to convert chemical energy to electrical energy. The chemical reaction involves the exchange of electrons is required for an electrochemical cell and such chemical reactions are called redox reactions.