Question

In an electrochemical cell___________
(a) Potential energy change into Kinetic energy
(b) Kinetic energy change into Potential energy
(c) Chemical energy change into Electrical energy
(d) Electrical energy change into Electrical energy

Answer 1

Hint: An electrochemical cell consists of a negative (anode) and a positive (cathode) electrode respectively which are immersed in an electrolyte. The oxidation occurs at the anode whereas the reduction occurs at the cathode, thus generating electric current.

Complete Step by Step Solution:
The above question can be explained by taking the example of a galvanic cell. A galvanic cell is also known as an electrochemical cell.

Notation of a galvanic cell:
In a galvanic cell, \[Cu\] undergoes oxidation to \[C{u^{2 + }}\] at the anode and \[A{g^ + }\] to \[Ag\] at the cathode.
In an electrochemical cell notation, the oxidation half-reaction is noted first followed by the reduction half-reaction.
Therefore, the oxidation half-reaction at the anode can be represented as below:
 \[Cu(s) \to C{u^{2 + }}(aq.) + 2{e^ - }\] … (1)

The reduction half-reaction at the cathode can be represented as below:
 \[A{g^ + }(aq.) + {e^ - } \to Ag(s)\] … (2)

Hence, the standard cell notation for a galvanic cell can be shown as below:
 \[Cu(s)|C{u^{2 + }}(aq.)||A{g^ + }_{}(aq.)|Ag(s)\] … (3)

Flow of current in galvanic cells: We all know that current always flows opposite to the flow of electrons. Therefore, the galvanic cell notation represents that loss of electron or oxidation occurs at the anode. Whereas gain of electron or reduction takes place at the cathode i.e., the electron is flowing from anode to cathode. Therefore, the current will flow opposite to the electron current (from anode to cathode) or on the other hand we can say that in the galvanic cell or electrochemical cell the chemical reaction generates the electric current.

The half-cell reaction of a galvanic cell can be represented as:
At Anode: \[Cu(s) \to C{u^{2 + }}(aq.) + 2{e^ - }\]
At cathode: \[A{g^ + }(aq.) + {e^ - } \to Ag(s)\]
From the cell reaction, it is also clear that option C will be the correct answer.

Note: According to the Nernst equation on increasing the temperature, the voltage of the galvanic cell will decrease i.e voltage of the galvanic cell is inversely proportional to the temperature.