'O' & 'A' Level Qn.
Copper can reduce nitric acid to NO2. Write a balanced equation
for this reaction.
Is it :
1) Cu +2HNO3 +2H+ --> Cu (2+) + 2NO2 +
2H20 OR
2) Cu +4H(+) + 2NO3(-) --> Cu(2+) +2NO2+ 2H20
Write a balanced equation for the reduction of nitric acid to
NH4+ by aluminium.
Suggest a reason why the reduction products of nitric acid are
different for each of these two metals.
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Solution :
It depends on whether concentrated (ie. pure liquid, no water
present) or aqueous acid is used. If concentrated, the HNO3 exists
as a covalent, unionized molecule. If aqueous, it dissociates and
ionizes to H+ and NO3-.
If concentrated, use equation #1. If aqueous, use equation #2.
If the exam question is ambiguous, give both with qualifications
(ie. explanations).
Regarding the QA test for nitrate(V) ions, the aluminium foil or
dervada's alloy will reduce nitrate(V) to ammonium cation, which
will undergo a Bronsted-Lowry acid-base proton transfer reaction
with the hydroxide ion (from the alkali added), to generate aqueous
ammonia, which when heated overcomes the hydrogen bonds with water
and generates gaseous ammonia, which can be detected by the
hydrolysis of ammonia on moist red litmus paper generating the
hydroxide ion, turning the litmus paper blue.
Reduction half-equation :
NO3(-) + 10H(+) + 8e- ---> NH4(+) + 3H2O
Oxidation half-equation :
Al ---> Al(3+) + 3e-
Combine the two half-equations to get the balanced redox
equation.
The reasons for the different redox products of nitric(V) acid
with aluminium versus copper, has to do with the different
oxidation potentials (ie. reducing strength) of Al versus Cu. Al is
a lot more reactive, has a more positive oxidation potential, and
is hence a more powerful reducing agent than Cu.
Hence Al is able to reduce the O.S. of N from +5 (in HNO3) all
the way to -3 (in NH4+), while Cu is only able to reduce the O.S.
of N from +5 (in HNO3) down slightly to +4 (in NO2).
