Question

In the electrochemical cell using the redox reaction below, the reduction half-reaction is ________.

Sn4+ (aq) + Fe (s) ? Sn2+ (aq) + Fe2+ (aq)
In the electrochemical cell using the redox reaction below, the reduction half-reaction is ________.
(aq) + (s) (aq) + (aq)

Fe+2e−→Sn2+
Fe→Fe2++2e−
Sn4+→Sn2++2e−
Fe+2e−→Fe2+
Sn4++2e−→Sn2+

Answer 1

Answer: In the given electrochemical cell, the reduction half reaction is
`Sn^(4+)(aq.)+2e^-\rightarrow Sn^(2+)(aq.)`

Step-by-step 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.

`X\rightarrow X^(n+)+ne^-`

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

`X^(n+)+ne^-\rightarrow X`

For the given chemical reaction:

`Sn^(4+)(aq.)+Fe(s)\rightarrow Sn^(2+)(aq.)+Fe^(2+)(aq.)`

Oxidation half reaction:
`Fe(s)\rightarrow Fe^(2+)(aq.)+2e^-`

Reduction half reaction:
`Sn^(4+)(aq.)+2e^-\rightarrow Sn^(2+)(aq.)`

Hence, in the given electrochemical cell, the reduction half reaction is
`Sn^(4+)(aq.)+2e^-\rightarrow Sn^(2+)(aq.)`