Question

the following skeletal oxidation - reduction reaction occurs under basic conditions. write the balanced oxidation half reaction.
s + bi → bi(oh)₃ + s²⁻

Explanation:

Step1: Determine oxidation states

S has an oxidation state of 0 in elemental S and - 2 in $S^{2-}$. Bi has an oxidation state of 0 in elemental Bi and + 3 in $Bi(OH)_3$. Oxidation is the loss of electrons. Here, Bi is oxidized.

Step2: Write the un - balanced oxidation half - reaction

$Bi
ightarrow Bi(OH)_3$

Step3: Balance elements other than O and H

Bi is already balanced on both sides.

Step4: Balance O by adding $OH^-$ in basic medium

$Bi + 3OH^-
ightarrow Bi(OH)_3$

Step5: Balance H by adding $H_2O$

$Bi + 3OH^-
ightarrow Bi(OH)_3$ (H is already balanced in this case)

Step6: Balance charge by adding electrons

$Bi + 3OH^-
ightarrow Bi(OH)_3+3e^-$

Answer:

$Bi + 3OH^-
ightarrow Bi(OH)_3+3e^-$