Posted by **Yukiko** on Wednesday, September 3, 2008 at 1:36am.

Balance the following redox equations. All occur in Acidic solutions.

1. Cr2O7^2-^ + C2O4^2-^ -> Cr^3+^ + CO2

2. Cu + NO3^-^ -> CU^2+^ + NO

3. MnO2+ HNO2 -> Mn^2+^ + NO3^-^

4. PbO2 + Mn^2+^ + SO4^2-^ -> PbSO4 + MnO4^-^

5. HNO2 + Cr2O7^2-^ -> Cr^3+^ + NO3^-^

- Chemistry -
**GK**, Wednesday, September 3, 2008 at 11:09am
Balancing oxidation-reduction (redox) reaction is a complex process. You must know how to **assign** some oxidation numbers and how to ** calculate** others no matter which method you use. The best method in my opinion is the “Ion-Electron” method. This is not a lesson on how to do it. Your textbook and your class notes should take care of that. It is only a demonstration using your first reaction:

A. Determine which element is reduced and which is oxidized.

1. Find out which element is reduced (Its oxidation number decreases). The oxidation number of Cr in Cr2O7^-2 is found by assigning -2 as the oxidation number of O, and x to Cr:

2x + (-2)(7) = -2 (the -2 on the right side is the ionic charge)

Solving for x we get x = +6. So Cr on the left side = +6 on the right side it is +3.

2. Find out which element is oxidized. The only possibility is C. Why? Looking at C2O4^-2, we can set up:

2y + (-2)(4) = -2, and y = 3.

B. Write incomplete half reaction for reduction and for oxidation, then complete them and balance them for number of atoms and electrical charge:

1. Reduction

Cr2O7^-2 —> Cr^+3

(balancing atoms and charge by adding H+, H2O, and e^- as needed)

Cr2O7^-2(aq) + 14H^+(aq) + 6e^- —> 2Cr^+3 + 7H2O

(check number of atoms of each kind and total charge on both sides)

2. Oxidation

C2O4^2- —> CO2

(balancing atoms and charge)

C2O4^2- —> 2CO2 + 2e^-

C. Rewrite and reconcile the two half reactions, then add them and simplify:

Cr2O7^-2(aq) + 14H^+(aq) + 6e^- —> 2Cr^+3 + 7H2O

3C2O4^2- —> 6CO2 + 6e^-

(I multiplied the oxidation by 3 so that the total number of electrons transferred will be 6 as it is in the reduction)

Cr2O7^-2(aq) + 14H^+(aq) + 6e^- + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2 + 6e^-

(added left and right sides ... then we simplify through cancellations)

Cr2O7^-2(aq) + 14H^+(aq) + 3C2O4^2- —> 2Cr^+3 + 7H2O + 6CO2

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