The balanced reaction is

\(K\left( s \right) + {H_2}O\left( l \right) \to KOH\left( l \right) + {H_2}\)

Here potassium is the oxidizing atom that has the highest oxidation state, i.e., +1.

The oxidation-reduction reaction is

Oxidation-\(K \to {K^ + } + 1{e^ - }\)

Reduction-\(2{H^ + } + 2{e^ - } \to {H_2}\)

Thus, oxidation of potassium and reduction of hydrogen takes place.

Oxidation number:

The oxidation state of K changes from 0 to +1,which means potassium loses 1 electron and gets oxidized,whereas hydrogen changes from +1 to 0,which means hydrogen gains 1 electron and gets reduced.

Here, potassium is the oxidizing atom andhas the highest oxidation state of +1.

The balanced reaction is

\(Ba\left( s \right) + 2HBr\left( {aq} \right) \to BaB{r_2} + {H_2}\)

Here barium is theoxidizing atom with a +2oxidation state.

The oxidation-reduction reaction is

Oxidation-\(Ba \to B{a^{2 + }} + 2{e^ - }\)

Reduction-\(2{H^ + } + 2{e^ - } \to {H_2}\)

Oxidation number:

Here,barium loses 2 electronsand changes from 0 to +2,showing oxidation,whereas hydrogen gains 1 electron and changes from +1 to 0,showingreduction.

So, barium is the oxidizing atomwith a+2oxidation state.

The balanced reaction is

\(Sn\left( s \right) + {I_2}\left( s \right) \to Sn{I_2}\)

Here, tinis the oxidizing atom with an oxidation state of +2.

The oxidation-reduction reaction is

Oxidation-\(Sn \to S{n^{2 + }} + 2{e^ - }\)

Reduction- \({I_2} + 2{e^ - } \to 2{I^ - }\)

Oxidation number:

In this reaction, Sn changes from 0 to +2,which means it loses 2 electrons and getsoxidized,whereas I change from 0 to -1,which means it gains 1 electron and getsreduced.

So, tin is the oxidizing atom with an oxidation state of +2.