Can some1 teach me how to balance redox using half equation
method?
EG: Cu + HNO3 -> Cu(NO3)2 + NO
I2 + NA2S2O3 -> NAI + NA2S4O6
Here is how you balance half-equations :
Hydrogen peroxide is reduced to water (chemical name : hydrogen
hydroxide aka dihydrogen monoxide aka hydrogen(I) oxide).
Hydrogen peroxide is oxidized to molecular oxygen gas.
(Acidified) Manganate(VII) ions are reduced to Manganese(II)
cations.
Nitrogen dioxide is oxidized to Nitrate(V) anions.
[reduction]
H2O2 --> H2O
To balance oxygen,
H2O2 --> 2H2O
To balance hydrogen,
2H+ + H2O2 --> 2H2O
To balance charges,
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O2 --> O2
To balance hydrogen,
H2O2 --> O2 + 2H+
To balance charges,
H2O2 --> O2 + 2H+ + 2e-
[reduction]
MnO4 - --> Mn2+
To balance oxygen,
MnO4 - --> Mn2+ + 4H2O
To balance hydrogen,
8H+ + MnO4 - --> Mn2+ + 4H2O
To balance charges,
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
[oxidation]
NO2 --> NO3 -
To balance oxygen,
H2O + NO2 --> NO3 -
To balance hydrogen,
H2O + NO2 --> NO3 - + 2H+
To balance charges,
H2O + NO2 --> NO3 - + 2H+ + e-
To obtain overall balanced redox equations, we ensure no. of
electrons lost = no. of electrons gained (by multiplying one or
both half-equations to obtain the lowest common multiple for
coefficient of electrons), then add up the half equations
(left hand side + left hand side, right hand side + right hand
side), and finally cancel away any common species on both sides,
invariably including all electrons.
For the reaction of using acidified KMnO4(aq) on H2O2, we
have
[reduction]
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
16H+ + 2MnO4 - + 10e- --> 2Mn2+ + 8H2O
[oxidation]
H2O2 --> O2 + 2H+ + 2e-
5H2O2 --> 5O2 + 10H+ + 10e-
[Balanced Redox]
(16H+ + 2MnO4 - + 10e-) + (5H2O2) --> (2Mn2+ + 8H2O) +
(5O2 + 10H+ + 10e-)
6H+ + 2MnO4 - + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
For the reaction of bubbling NO2 into H2O2, we have
[reduction]
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O + NO2 --> NO3 - + 2H+ + e-
2H2O + 2NO2 --> 2NO3 - + 4H+ + 2e-
[Balanced Redox]
(2H2O + 2NO2) + (2H+ + 2e- + H2O2) --> (2NO3 - + 4H+ + 2e-) +
(2H2O)
2NO2 + H2O2 --> 2NO3 - + 2H+
The Question :
300 cm3 of a mixture of
dinitrogen monoxide and nitrogen dioxide, at r.t.p. conditions is
bubbled through 75 cm3 of 0.10 mol/dm3 acidified hydrogen
peroxide solution. The nitrogen dioxide is oxidized to nitrate(V)
ions, while the inert dinitrogen monoxide does not react. The
remaining hydrogen peroxide in 50.0 cm3 of the resulting solution
is then titrated with a 0.050 mol/dm3 acidified potassium
manganate(VII) solution, of which 22.0 cm3 was required. Calculate
the % by volume of dinitrogen monoxide in the gas
mixture.
Final Answer :
46%.
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Here's a useful list of redox half equations.
Note that you can obtain the oxidation potential by changing the
-ve/+ve sign. For instance,
The standard reduction potential for Ag+ to Ag is +0.8V.
The standard oxidation potential for Ag to Ag+ is -0.8V.
The standard reduction potential for Zn2+ to Zn is -0.76V.
The standard oxidation potential for Zn to Zn2+ is +0.76V.
