taylorr6
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Can someone help me solve this? I was given the answer at the end of class but I don't understand how the teacher got it. (Directions: Balance each redox reaction in acid solution using the half reaction method.)
PbO2+I2=>Pb^2+ +IO3^- = I2+8H^+ +5PbO2=>2IO3^- +5Pb^+2 +4H2O

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Unbalanced chemical reaction: PbO₂ + I₂ → Pb²⁺ +IO₃⁻.
First determine oxidation numbers of elements:
1) lead on the left side of reaction has oxidation number +4 (x + 2·(-2) = 0) and on the right side +2, so lead is reduced.
2) iodine has neutal charge (0) on left and oxidation number +5 on the right side of chemical reaction (x + 3 · (-2) = -1), so iodine is oxidized.
3) oxygen has oxidation number -2 on both side of chemical reaction.
Second, write balanced half reactions, because this reaction is acidic solution, add hydrogen ions on one side and water on another side of half reaction; balance atoms and electrons on both half reactions:
Half reaction 1: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O /×5
5PbO₂ + 20H⁺ + 10e⁻ → 5Pb²⁺ + 10H₂O.
Half reaction 2: I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻.
Sum both half reactions and short same ions: 
5PbO₂ + 20H⁺ + I₂ + 6H₂O → 5Pb²⁺ + 10H₂O + 2IO₃⁻ + 12H⁺.
Abbreviate reaction: 5PbO₂ + 8H⁺ + I₂  → 5Pb²⁺ + 4H₂O + 2IO₃⁻..
mahima6824soni

The balanced equation is [tex]\rm 5PbO_2 + 8H^+ + I_2 = 5Pb^2^+ + 4H_2O + 2IO_3[/tex]

What is a redox reaction?

A redox reaction is also known as an oxidation-reduction reaction.

In this type of reaction, there is the transfer of electrons between two species.

The balanced equation is

[tex]\rm PbO_2 + I_2 = Pb^2^+ +IO_3^-[/tex]

Determine the oxidation number of elements

The reactant's oxidation number will be +4 (x + 2·(-2))= 0

The product's oxidation number is +2

Now, iodine has a neutral charge (0) and its oxidation number is +5, thus iodine is oxidized.

The oxidation number of oxygen is -2 on both side

The equation for balanced first half-reaction is

[tex]PbO_2 + 4H^+ + 2e^- = Pb^2^+ + 2H_2O /\times 5[/tex]

[tex]\rm 5PbO_2 + 20H^+ + 10e^- = 5Pb^2^+ + 10H_2O.[/tex]

The second half-reaction

[tex]\rm I_2+ 6H_2O = 2IO_3^- + 12H^+ + 10e^-\\[/tex]

Add both reactions

[tex]\rm 5PbO_2 + 20H^+ + I_2 + 6H_2O = 5Pb^2^+ + 10H_2O + 2IO_3^- + 12H^+[/tex]

Thus, the balanced equation is [tex]\rm 5PbO_2 + 8H^+ + I_2 = 5Pb^2^+ + 4H_2O + 2IO_3[/tex]

Learn more about redox, here:

https://brainly.com/question/13978139

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