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

Answer:

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

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.