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: $I_2\rightarrow IO_3^-$ Reduction half - reaction: $PbO_2\rightarrow Pb^{2 + }$

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

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

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

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

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

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

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

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

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

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

Step7: Convert to basic medium

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

Answer:

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