# Redox Reaction

In an acidic solution, manganese(II) ion is oxidized to permanganate ion by bismuthate ion, BiO3-. In the reaction, BiO3- is reduced to Bi3+.

How many milligrams of NaBiO3 are needed to oxidize the manganese in 30.2 mg of manganese(II) sulfate?

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1. Write the redox reaction:
Mn2+ + BiO3- --→ Bi3+ + MnO4−

Balance this redox reaction. Write first the half-reactions:
Reduction: BiO3- ---> Bi3+ (charge is changed from 5+ to 3+)
Oxidation: Mn2+ ---> MnO4- (charge is changed from 2+ to 7+)

Balance them by adding H+ (acidic media), H2O and e- :
Reduction: 2e- + 6H+ + BiO3- ---> Bi3+ + 3 H2O
Oxidation: 4 H2O + Mn2+ ---> MnO4- + 8H+ + 5e-

Multiply each half-reaction by some number to make their number of electrons equal, so when added will cancel them out.
5 * ( 2e- + 6H+ + BiO3- ---> Bi3+ + 3 H2O )
2 * ( 4 H2O + Mn2+ ---> MnO4- + 8H+ + 5e- )
---------------------------------------------------------
14 H+ + 5 BiO3^- + 2 Mn2+ ---> 5 Bi3+ + 7 H2O + 2 MnO4-

Now that you have the balanced reaction, you can solve for the problem. Steps to solve:
(1) Find the molar mass of Manganese(II) Sulfate (MnSO4).
(2) Divide the given mass by the molar mass of MnSO4.
(3) Make a mole ratio of BiO3- : Mn2+ from the balanced reaction.
(4) Multiply the answer you got from #2 by #3. This answer is the moles of BiO3-.
(5) Finally, multiply the answer in #4 by the molar mass of NaBiO3.

hope this helps~ `u`

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2. Step1: assign oxidation numbers.

Cr2O72- + C2O42- → 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).

Oxidation:

C2O42- →CO2

Reduction:

Cr2O72- → Cr3+

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

First, balance all elements other than Hydrogen and Oxygen.
C2O42- →2CO2

Cr2O72- → 2Cr3+

Second, balance Oxygen by adding H2O.
C2O42- →2CO2

Cr2O72- → 2Cr3+ + 7H2O

Third, balance Hydrogen by adding H+.
C2O42- →2CO2

14H+ + Cr2O72- → 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- →2CO2 + 2e–

6e– +14H+ + Cr2O72- → 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- →2CO2 + 2e–] = 3C2O42- →6CO2 + 6e–

6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O

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

3C2O42- →6CO2 + 6e–

+

6e– +14H+ + Cr2O72- → 2Cr3+ + 7H2O

3C2O42- +14H+ + Cr2O72- → 2Cr3+ + 7H2O + 6CO2

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

Hope this helps.

-CoolChemTutor

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