Question

homework: balance the redox reactions in both acids basic media ② $pbo_2 + i_2 longrightarrow pb^{2+}+io_3^-$

Explanation:

Step1: Determine oxidation - reduction half - reactions

Oxidation half - reaction: $I_2
ightarrow IO_3^-$
Reduction half - reaction: $PbO_2
ightarrow Pb^{2 + }$

Step2: Balance atoms other than O and H in acidic medium

Oxidation: $I_2
ightarrow 2IO_3^-$
Reduction: $PbO_2
ightarrow Pb^{2 + }$

Step3: Balance O by adding $H_2O$ in acidic medium

Oxidation: $I_2 + 6H_2O
ightarrow 2IO_3^-$
Reduction: $PbO_2
ightarrow Pb^{2 + }+2H_2O$

Step4: Balance H by adding $H^+$ in acidic medium

Oxidation: $I_2 + 6H_2O
ightarrow 2IO_3^-+12H^+$
Reduction: $PbO_2 + 4H^+
ightarrow Pb^{2 + }+2H_2O$

Step5: Balance charge by adding $e^-$ in acidic medium

Oxidation: $I_2 + 6H_2O
ightarrow 2IO_3^-+12H^++10e^-$
Reduction: $PbO_2 + 4H^++2e^-
ightarrow Pb^{2 + }+2H_2O$

Step6: Multiply half - reactions to equalize electrons and add them in acidic medium

Multiply reduction half - reaction by 5.
Oxidation: $I_2 + 6H_2O
ightarrow 2IO_3^-+12H^++10e^-$
Reduction: $5PbO_2 + 20H^++10e^-
ightarrow 5Pb^{2 + }+10H_2O$
Overall in acidic medium: $5PbO_2+I_2 + 8H^+
ightarrow 5Pb^{2 + }+2IO_3^-+4H_2O$

Step7: Convert to basic medium

Add $OH^-$ to both sides to neutralize $H^+$
$5PbO_2+I_2 + 8H^++8OH^-
ightarrow 5Pb^{2 + }+2IO_3^-+4H_2O+8OH^-$
Since $H^++OH^- = H_2O$, we get:
$5PbO_2+I_2 + 4H_2O
ightarrow 5Pb^{2 + }+2IO_3^-+8OH^-$

Answer:

Acidic medium: $5PbO_2+I_2 + 8H^+
ightarrow 5Pb^{2 + }+2IO_3^-+4H_2O$
Basic medium: $5PbO_2+I_2 + 4H_2O
ightarrow 5Pb^{2 + }+2IO_3^-+8OH^-$