Question

In a Daniel Cell:
$(a){\text{ }}$The chemical energy liberated during the redox reaction is converted to electrical energy.
$(b){\text{ }}$The electrical energy of the cell is converted to chemical energy.
$(c){\text{ }}$The energy of the cell is utilized in conduction of the redox reaction.
$(d){\text{ }}$The potential energy of the cell is converted into the electrical energy.

Answer 1

Hint: The basic structure of the Daniel cell must be known. The chemical reactions occurring at the respective anode and cathode leads to the production of the electric current. The salt bridge plays an important role by allowing anions to the anode compartment and cations to the cathode compartments.

Complete step by step solution:
A Daniel cell is a galvanic cell which converts the chemical energy into the electrical energy. The Daniel cell is so designed that it produces electric current by spontaneous redox reaction between zinc and cupric ion.
The construction of the Daniel cell includes a copper container which is filled with a concentration solution of copper sulphate. A porous cylindrical pot filled with the diluted sulphuric acid is immersed in the copper container. One amalgamated zinc rod is immersed in the diluted sulphuric acid. The two different metals are connected to each other via a salt bridge, which maintains the electrical neutrality by preventing the accumulation of the charges by maintaining a free path for the migration of ions.
Cell reactions:
At anode, in the $Zn/ZnS{O_4} $ half cell, the oxidation reaction occurs.
$Zn(s) \to Z {n^{2 + }}(aq) + 2{e^ - }$
At cathode, ion $Cu/CuS{O_4} $ half cell, the reduction reaction occurs.
$C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)$
The net cell reaction is
$Zn(s) + C{u^{2 + }}(aq) \rightleftharpoons Z{n^{2 + }}(aq) + Cu(s)$
Therefore, we have seen that in the above reaction, the chemical energy is liberated when the oxidation at the anode end and the reduction at the cathode end takes place in the Daniel cell and it is then converted into the electrical energy.
Hence, the correct solution is the option$(a)$.

Note: The students might get confused about where the reduction and oxidation take place. So, in order to memorise this, remember the mnemonic term “REDCAT” i.e., Reduction (RED) occurs at the cathode (CAT), and oxidation occurs at the anode.