How do you solve this pls help?

Question

H2C2O4 + MnO4¯ ---> CO2 + Mn2+

1) The balanced half-reactions:

2) Equalize the electrons:

3) The final answer

O2 + As ---> HAsO2 + H2O

1) correct half-reactions:

2) equalize the electrons

3) The final answer:

Answer 1

You need to know the oxidation number rules and how to calculate them so that you know what is oxidized and what is reduced.

C is oxidized from +3 to +4 state and Mn is reduced from +7 to +2 state

Before you balance the half-reactions, you need to know whether the conditions are acidic or basic.

1) Write skeleton equation and balance the species undergoing Ox or Red'n: (In this case C)

H₂C₂O₄ → 2CO₂

The way that's usually taught:

2) You use H₂O to balance O, which is unnecessary here.

3) Use H⁺ to balance H

H₂C₂O₄ → 2CO₂ + 2H⁺

4) Use e^- to balance overall charge

H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e^-

Do the same for Mn:

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Another process (that I prefer)

2) Use oxidation states to determine electrons needed

3) Use H⁺ or OH^- to balance charge

4) Use H₂O to balance H or O and the other must also work (It's a nice check)

In this method, you know that you know 5 e^- to get Mn from +7 to +2, you need 8H⁺ in order to balance the charge, you need 4H₂O to balance the O (and the H's are also balanced)

In order to combine the 1/2 - rxns, you make it so that the number of electrons are the same (5 x 1st rxn + 2 x 2nd reaction) which yields

5H₂C₂O₄ + 2MnO₄^- + 6H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O

I'll leave other problems to you.

Answer 2

H₂C₂O₄ + MnO₄¯ ---> CO₂ + Mn²⁺ 

H₂C₂O₄ ==> CO₂ ... oxidation half reaction

MnO₄¯ ==> Mn²⁺ ... reduction half reaction

1) 

balancing oxidation half reaction:

H₂C₂O₄ ==> 2CO₂ + 2H⁺

balancing the reduction half reaction:

MnO₄¯ + 8H⁺ ==> Mn²⁺ + 4H₂O

2)

equalizing electrons:

H₂C₂O₄ ==> 2CO₂ + 2H⁺ + 2e- [x5] to use 10 electrons = 5H₂C₂O₄ ==> 10CO₂ + 10H⁺ + 10e-

MnO₄¯ + 8H⁺ + 5e- ==> Mn²⁺ + 4H₂O [x2] to use 10 electrons = 2MnO₄¯ + 16H⁺ + 105e- ==> 2Mn²⁺ + 8H₂O

3) final answer:

5H₂C₂O₄ + 2MnO₄¯ + 6H+ ==> 10CO₂ + 2Mn²⁺ + 8H₂O ... BALANCED REDOX EQUATION

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O₂ + As ---> HAsO₂ + H₂O

As ==> HAsO2 ... oxidation half reaction

O2 ===> H2O ... reduction half reaction

1) balancing half reactions

As + 2H₂O ==> HAsO₂ + 3H⁺ 

O₂ + 4H⁺ ==> 2H₂O 

2) equalize electrons:

As + 2H₂O ==> HAsO₂ + 3H⁺ + 3e- [x4] = 4As + 8H₂O ==> 4HAsO₂ + 12H⁺ + 12e-

O₂ + 4H⁺ + 4e- ==> 2H₂O [x3] = 3O₂ + 12H⁺ + 12e- ==> 6H₂O

3) final answer:

4As + 8H₂O + 3O₂ ==> 4HAsO₂ + 6H₂O ... BALANCED REDOX EQUATION