Answer: In the given electrochemical cell, the reduction half reaction is [tex]Sn^{4+}(aq.)+2e^-\rightarrow Sn^{2+}(aq.)[/tex]
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]Sn^{4+}(aq.)+Fe(s)\rightarrow Sn^{2+}(aq.)+Fe^{2+}(aq.)[/tex]
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]Sn^{4+}(aq.)+2e^-\rightarrow Sn^{2+}(aq.)[/tex]
Hence, in the given electrochemical cell, the reduction half reaction is [tex]Sn^{4+}(aq.)+2e^-\rightarrow Sn^{2+}(aq.)[/tex]
