Question

What is the cell reaction occurring in a Daniel cell (galvanic cell)?
(A) $Cu(s)+ZnS{{O}_{4}}(aq)\to CuS{{O}_{4}}(aq)+Zn(s)$
(B) $CuS{{O}_{4}}(aq)+Zn(s)\to Cu(s)+ZnS{{O}_{4}}(aq)$
(C) $Ni(s)+ZnS{{O}_{4}}(aq)\to NiS{{O}_{4}}(aq)+Zn(s)$
(D) $2Na(s)+CdS{{O}_{4}}(aq)\to N{{a}_{2}}S{{O}_{4}}(aq)+Cd(s)$

Answer 1

A Daniel cell or galvanic cell is a device that can generate electric energy from chemical energy. Electrical energy brings out a chemical reaction with the help of an external energy source. Both the anode and cathode are in the same container with an electrolyte solution

Complete step by step answer:
Daniel cell with two different metal electrodes and each electrode is in contact with its own solution.
When an electrode dipped in its own solution, the potential developed, is known as standard potential. If the process is at anode electrode, it is standard oxidation potential, vice versa for reduction potential electrode.
The construction of these two half cells form a cell which generates EMF of a cell. In the two half cell, one is anode where oxidation takes place and another half cell Is cathode where reduction takes place.
The total EMF of cell ${{E}_{cell}}^{o}={{E}_{oxd}}^{o}+{{E}_{red}}^{o}$

(A) $Cu(s)+ZnS{{O}_{4}}(aq)\to CuS{{O}_{4}}(aq)+Zn(s)$
In this cell reaction, the standard potential of $Zn/Z{{n}^{+2}}=0.76V\And C{{u}^{+2}}/Cu=0.34V$
The given cell reaction is not feasible because the emf of this cell is -1.1V. If the emf value is negative, the cell construction is not possible.

(B) $CuS{{O}_{4}}(aq)+Zn(s)\to Cu(s)+ZnS{{O}_{4}}(aq)$
$Zn(s)$ Oxidizes to $ZnS{{O}_{4}}(aq)$
$CuS{{O}_{4}}(aq)$ Reduces to $Cu(s)$
The emf of the cell is +1.1V. hence the cell reaction is spontaneous. Hence this is the Daniel cell reaction.

(C) $Ni(s)+ZnS{{O}_{4}}(aq)\to NiS{{O}_{4}}(aq)+Zn(s)$
Due to the high oxidation potential of Zn, it does not reduce. This cell reaction is not feasible.

(D) $2Na(s)+CdS{{O}_{4}}(aq)\to N{{a}_{2}}S{{O}_{4}}(aq)+Cd(s)$
This cell reaction is also not spontaneous due to potential energy differences.
Hence, the correct cell reaction occurring in Daniel cell is $CuS{{O}_{4}}(aq)+Zn(s)\to Cu(s)+ZnS{{O}_{4}}(aq)$
So, the correct answer is “Option B”.

Note: An electrolytic cell is a type of electrochemical cell in which electrical energy is converted to chemical energy. Chemical energy generates electrical energy with the help of a redox reaction. Both the anode and cathode are in different containers known as a half cell, which are connected with a salt bridge.