Balance the redox equation by the ion-electron method: Bi(OH)_3+SnO_2^{2-} to SnO_3^{2-}+ Bi in...

Let's first deduce the half-reaction for the oxidation.

{eq}\rm SnO_2^{2-}\to SnO_3^{2-} {/eq}

Balance the oxygen atoms using {eq}\rm H_2O {/eq}.

{eq}\rm SnO_2^{2-}+H_2O \to SnO_3^{2-} {/eq}

Balance the hydrogen atoms using {eq}\rm H^+ {/eq}.

{eq}\rm SnO_2^{2-}+H_2O \to SnO_3^{2-}+2H^+ {/eq}

Neutralize the hydrogen ions using {eq}\rm OH^- {/eq}:

{eq}\rm SnO_2^{2-}+H_2O ++2OH^- \to SnO_3^{2-}+2H^+ +2OH^-\\ \rm SnO_2^{2-}+H_2O ++2OH^- \to SnO_3^{2-}+2H_2O\\ {/eq}

Balance the charge:

{eq}\rm SnO_2^{2-}+H_2O ++2OH^- \to SnO_3^{2-}+2H_2O+2e^-\ \ \ (1) {/eq}

Let's repeat the same for the reduction reaction:

{eq}\rm Bi(OH)_3 \to Bi {/eq}

Here, we can just balance the reaction directly using {eq}\rm OH^- {/eq}:

{eq}\rm Bi(OH)_3 \to Bi+3OH^- {/eq}

Balance the charge:

{eq}\rm Bi(OH)_3 +3e^-\to Bi+3OH^-\ \ \ (2) {/eq}

Match the number of electrons in the half-reactions. The lowest common multiple for the number of electrons is 6.

Equation (1) multiplied by 3, we get:

{eq}\rm 3SnO_2^{2-}+3H_2O ++6OH^- \to 3SnO_3^{2-}+6H_2O+6e^-\ \ \ (3) {/eq}

Equation (2) multiplied by 2, we get:

{eq}\rm 2Bi(OH)_3 +6e^-\to 2Bi+6OH^-\ \ \ (4) {/eq}

Combine (3) and (4):

{eq}\begin{align*} \rm 3SnO_2^{2-}+3H_2O +6OH^- + 2Bi(OH)_3 +6e^- \to 3SnO_3^{2-}+6H_2O+6e^-+2Bi+6OH^-\\ \rm 3SnO_2^{2-} + 2Bi(OH)_3 \to 3SnO_3^{2-}+(6-3)H_2O+2Bi+(6-6)OH^-\\ \color{blue}{\boxed{\rm 3SnO_2^{2-} + 2Bi(OH)_3 \to 3SnO_3^{2-}+3H_2O+2Bi}} \end{align*} {/eq}