Question

When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)3– will be _____. Bi(OH)3(s) + Sn(OH)3–(aq) → Sn(OH)62–(aq) + Bi(s) (basic solution)

Answer 1

The coefficient of
`Sn(OH)_3^(-)` is 3 in the balanced redox reaction.

Step-by-step explanation:

Oxidation reaction is defined as the chemical reaction in which an atom loses its electrons. The oxidation number of the atom gets increased during this reaction.

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

Reduction reaction is defined as the chemical 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:

`Bi(OH)_3+Sn(OH)_3^(-)\rightarrow Sn(OH)6^(2-)+Bi`

The half cell reactions for the above reaction follows:

Oxidation half reaction:
`Sn(OH)_3^(-)+3OH^-\rightarrow Sn(OH)6^(2-)+2e^-`

Reduction half reaction:
`Bi(OH)_3+3e^-\rightarrow Bi+3OH^-`

To balance the oxidation half reaction must be multiplied by 3 and reduction half reaction must be multiplied by 2 thus, the balanced equation is:-

`3Sn(OH)_3^(-)+3OH^-+2Bi(OH)_3\rightarrow 3Sn(OH)6^(2-)+2Bi`

The coefficient of
`Sn(OH)_3^(-)` is 3 in the balanced redox reaction.