Respuesta :
Explanation:
The given cell reaction is as follows.
[tex]In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)[/tex]
Hence, reactions taking place at the cathode and anode are as follows.
At anode ; Oxidation-half reaction : [tex]In(s) \rightarrow In^{3+}(aq) + 3e^{-}[/tex] ...... (1)
At cathode; Reduction-half reaction : [tex]Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)[/tex] ....... (2)
Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.
Therefore, net cell reaction is as follows.
[tex]2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}[/tex]
[tex]3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)[/tex]
Net reaction: [tex]2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)[/tex]
Thus, we can conclude that the overall cell reaction is as follows.
[tex]2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)[/tex]
The net ionic equation of the reaction occutung in the electrochemical cell involves oxidation of indium and reduction of cadmium ion.
What electrochemical cells?
Electrochemical cells are cells which produces electrical energy from chemical reactions occurring within the cells.
The given cell reaction of the electrochemical cell is as follows.
- In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)
The half reactions taking place at the cathode and anode are given below:
At anode; Oxidation-half reaction:
- In(s) → In3+ (aq) +3e-
At cathode; Reduction-half reaction:
Cd2+ (aq) +2e- → Cd(s)
The half reactions are balanced by multiplying equation (1) by 2 and equation (2) by 3.
- 2 In(s) → 2 In3+ (aq) +6e-
- 3 Cd2+ (aq) + 6e- → 3 Cd(s)
The net cell reaction is given as:
2 In(s) + 3 Cd2+ (aq) → 2 In3+ (aq) + 3 Cd(s)
Therefore, the net ionic equation of the reaction involves oxidation of indium and reduction of cadmium ion.
Learn more about net ionic equations at: https://brainly.com/question/25604204
