Question

Consider the reaction: $C{l_2} + 2KI \to {I_2} + 2KCl$
Which species will be reduced at the cathode?
A.$C{l_2}$
B.${I_2}$
C.${I^ - }$
D.$C{l^ - }$

Answer 1

Hint:Electrochemical reactions are those reactions which consume or produce free electrons. These reactions are redox reactions, i.e., reactions involving both reduction and oxidation. The number of electrons being exchanged is equal in an electrochemical reaction.

Complete answer:
Electrochemical cell consists of two electrodes, a cathode and an anode, and an electrolyte. Electrodes are conductive metal strips which are dipped in an electrolyte. Electrolyte is a compound which dissociates completely into the constituent positive and negative species, that is, cations and anions respectively.
The two electrodes are joined using a wire to complete the circuit and facilitate the flow of electrons. In this circuit, the cathode is the positively charged electrode and the anode is the negatively charged electrode. So, in the solution, we can say that the electrons flow towards the cathode and move away from the anode.
In the given reaction, $C{l_2} + 2KI \to {I_2} + 2KCl$
Iodine goes from an oxidation state of $ - 1$ to $0$ and chlorine goes from an oxidation state of $0$ to $ - 1$. We can see that chlorine accepts an electron over the course of the reaction and is getting reduced. Thus, the species that is reduced at the cathode is chlorine gas, $C{l_2}$.

Thus, the correct answer is A.

Note:
The loss of electrons from any species signifies that it is undergoing oxidation and the gain of electrons signifies the reduction of the species. Since electrons preferably are attracted towards the cathode, thus the species at the cathode will gain electrons and end up getting reduced. Similarly, since electrons move away from the anode, thus, the species at the anode will lose electrons and end up getting oxidized.