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^-^

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asked by Yukiko
  1. 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 β€œIon-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 β€”> Cr^+3
    (balancing atoms and charge by adding H+, H2O, and e^- as needed)
    Cr2O7^-2(aq) + 14H^+(aq) + 6e^- β€”> 2Cr^+3 + 7H2O
    (check number of atoms of each kind and total charge on both sides)
    2. Oxidation
    C2O4^2- β€”> CO2
    (balancing atoms and charge)
    C2O4^2- β€”> 2CO2 + 2e^-
    C. Rewrite and reconcile the two half reactions, then add them and simplify:
    Cr2O7^-2(aq) + 14H^+(aq) + 6e^- β€”> 2Cr^+3 + 7H2O
    3C2O4^2- β€”> 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- β€”> 2Cr^+3 + 7H2O + 6CO2 + 6e^-
    (added left and right sides ... then we simplify through cancellations)
    Cr2O7^-2(aq) + 14H^+(aq) + 3C2O4^2- β€”> 2Cr^+3 + 7H2O + 6CO2

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    posted by GK
  2. Step1: assign oxidation numbers.

    Cr2O72- + C2O42- β†’ Cr3+ + CO2

    +6/-2 +3/-2 +3 +4/-2

    Step2: Separate the overall reaction into two separate half reactions. One will be the oxidation reaction(where the oxidation number increased) and the other will be the reduction reaction( where the oxidation numbers decreased).


    C2O42- β†’CO2


    Cr2O72- β†’ Cr3+

    Step 3: Balance each half-reaction in the following order:

    First, balance all elements other than Hydrogen and Oxygen.
    C2O42- β†’2CO2

    Cr2O72- β†’ 2Cr3+

    Second, balance Oxygen by adding H2O.
    C2O42- β†’2CO2

    Cr2O72- β†’ 2Cr3+ + 7H2O

    Third, balance Hydrogen by adding H+.
    C2O42- β†’2CO2

    14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O

    Step 4: balance each half reaction with respect to charge by adding electrons. The sum of the charges on both sides of the equation should be equal.

    C2O42- β†’2CO2 + 2e–

    6e– +14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O

    Step 5: Make sure the number of electrons in both half reactions equal by multiplying one or both half reactions by a small whole number.

    3 x [C2O42- β†’2CO2 + 2e–] = 3C2O42- β†’6CO2 + 6e–

    6e– +14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O

    Step 6: Add the two half reactions together canceling electrons and other species as necessary.

    3C2O42- β†’6CO2 + 6e–


    6e– +14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O

    3C2O42- +14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O + 6CO2

    Final Answer: 3C2O42- +14H+ + Cr2O72- β†’ 2Cr3+ + 7H2O + 6CO2

    Hope this helps.


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    posted by CoolChemTutor
  3. C2o4- co3+

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    posted by Akshay kumar
  4. Cr2o72- + 3c2o42- +14H+ ----- 2cr3+ + 6co2 + the correct answer!!

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    posted by Tanmay kumar

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