In the following reactions you need to give both balanced half reactions in the

Question

In the following reactions you need to give both balanced half reactions in the appropriate

acidic or basic solution (one for oxidation and one for reduction) and then the overall balanced

equation. Do not break apart any of the following ions or molecules.

Balance in acidic solution:

1. MnO4+ H2C2O4 ? Mn+ CO2

2. As2S3`+ ClO3 ? H2AsO4+ SO4+ Cl

3. Cr(CN)6 + Ce ? Cr + Ce+ NO3+CO2

4. P4S3+ NO3 ? H3PO4+ SO4+ NO

5. As4O6+ Mn4? AsO4 + Mn

Balance in basic solution:

6. CrO2+ ClO ? CrO4 + Cl2

7. Cl2 + CrI3 ? Cl + CrO4+ IO4

8. C3H8O3 + V5O14 ? HCO2 + VO(OH)2

9. Ce + Fe(SCN)6 ? Ce(OH)3 + Fe(OH)3+ SO4+ CO3+ NO3

10. Al + NO2 ? AlO2+ NH3

Explanation / Answer

here's the steps...

1) write down an unbalanced equation showing all species
2) identify theoxidation stateof all atoms
3) determine which atoms were oxidized and which were reduced
4) write half reactions. include electrons
5) balance electrons in half reactions
6) combine half reactions and cancel e's
7) add counter ions and simplify
8) add any remaining species. THEN balance the remaining.

so let's apply those steps here...

*** 1 ***
__H2C2O4+ __ MnO4(-) + H(+) ---> __ Mn(+2) + __ CO2 + __H2O

*** 2 ***
H in H2C2O4 is +1
C in H2C2O4 is +3
O in O4 is -2
Mn in MnO4 is +7
O in MnO4 is -2
H in H+ is +1

Mn in Mn(+2) is +2
C in CO2 is +4
O in CO2 is -2
H in H2O is +1
O in H2O is -2

*** 3 ***
so...

all the Mn went from +7 to +2 and was "reduced".. reduction means reduction in oxidation state... ie reduction in charge.

and all of the C went from +3 to +4 and was "oxidized". went up in oxidation #.

the H's and O's remained unchanged.

*** 4 ***
Mn(+7) + 5 e's ---> Mn(+2)
C(+3) ---> C(+4) + 1 e

*** 5 ***
balance e's by multiplying the second by 5

Mn(+7) + 5 e's ---> Mn(+2)
5 C(+3) ---> 5 C(+4) + 5 e

*** 6 ***
Mn(+7) + 5 e's + 5 C(+3) ---> Mn(+2) + 5 C(+4) + 5 e

Mn(+7) + 5 C(+3) ---> Mn(+2) + 5 C(+4)

*** 7 ***
MnO4(-1) + 5/2 H2C2O4 ---> Mn(+2) + 5 CO2

2 MnO4(-1) + 5 H2C2O4 ---> 2 Mn(+2) + 10 CO2

*** 8 ***
adding remaining species...
2 MnO4(-1) + 5 H2C2O4 + ___ H+ ---> 2 Mn(+2) + 10 CO2 + ___ H2O

balancing O's...
2 MnO4(-1) + 5 H2C2O4 + ___ H+ ---> 2 Mn(+2) + 10 CO2 + 8 H2O

balancing H's...
2 MnO4(-1) + 5 H2C2O4 + 6 H+ ---> 2 Mn(+2) + 10 CO2 + 8 H2O

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