Complete and balance the following equations, and identify the oxidizing and red

Question

Complete and balance the following equations, and identify the oxidizing and reducing agents.(Express your answer as a chemical equation. Identify all of the phases in your answer.)

A. Cr2O2?7(aq)+I?(aq)?Cr3+(aq)+IO3-(aq) (acidic solution)

B. MnO?4(aq)+CH3OH(aq)?Mn2+(aq)+HCO2H(aq) (acidic solution)

C.I2(s)+OCl?(aq)?IO?3(aq)+Cl?(aq) (acidic solution)

D. As2O3(s)+NO?3(aq)?H3AsO4(aq)+N2O3(aq) (acidic solution)

E. MnO?4(aq)+Br?(aq)?MnO2(s)+BrO?3(aq) (basic solution)

F. Pb(OH)2?4(aq)+ClO?(aq)?PbO2(s)+Cl?(aq) (basic solution)

Explanation / Answer

A.

Cr2O72-(aq) + I? (aq) ? Cr3+ (aq) + IO3-(aq) (acidic solution)

Oxidising agent = Cr2O72- because its oxdidation number is decreased from + 6 to +3

Reducing agent = I- because its oxidation number is increased fro - 1 to + 5

1. Separate the redox reaction into oxidation half and reduction half reactions.

Oxidation Half Reaction        Reduction Half Reaction

I- -------> IO3-                            Cr2O72- -----------> Cr3+

2. Balance all atoms other than O and H

I- -------> IO3-                            Cr2O72- -----------> 2 Cr3+

3. To balance oxygen atoms in acidic medium add H2O molecules.

I- + 3 H2O -------> IO3-                            Cr2O72- -----------> 2 Cr3+ + 7 H2O

4. To balance hydrogen atoms in acidic medium add H+ ions.

I- + 3 H2O -------> IO3- + 6 H+          Cr2O72- + 14 H+ -----------> 2 Cr3+ + 7 H2O

5. To balance charges add electrons at more positive charge side

I- + 3 H2O -------> IO3- + 6 H+ + 6 e-    Cr2O72- + 14 H+ + 6 e- -----------> 2 Cr3+ + 7 H2O

6. To equalise electrons in both half reaction, do OHR * 1 and RHR * 1

I- + 3 H2O -------> IO3- + 6 H+ + 6 e-    Cr2O72- + 14 H+ + 6 e- -----------> 2 Cr3+ + 7 H2O

7. Now, add the balanced half reactions to get balanced redox reaction.

I- + 3 H2O -------> IO3- + 6 H+ + 6 e-  

Cr2O72- + 14 H+ + 6 e- -----------> 2 Cr3+ + 7 H2O

------------------------------------------------------------------------

I- + Cr2O72- + 8 H+ ----------> 2 Cr3+ + IO3- + 4 H2O

-------------------------------------------------------------------------

Verify,

Atoms Left Right

I            1        1

Cr         2         2

O          7 7

H          8         8

Charge + 5    + 5

Hence balanced.

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