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Electrochemistry II: Balancing Equations
are commonly used in the laboratory because they are so soluble and readily
available, and because they are not appreciably oxidizing in neutral aqueous
solution.
We have already seen that strongly acid solutions will make the oxygencontaining negative ions even stronger oxidizing agents; the nitrate ion in an
acid solution is no exception. When concentrated HNO 3 is used, the half-cell
potential is comparable to that of MnOj, but the gaseous product is no longer
NO; it is NO 2 . For concentrated HNO 3 , the half-reaction that must be used is
e- + 2H+ + NOJ ** H 2 O + NO 2
It is a common error to think that a metal, if it reacts with an acid, will produce
H 2 gas. This is not true if the acid is HNO 3 ; the gaseous products will be NO or
NO 2 , depending on the concentration. Under certain circumstances (usually
involving dilute solution and a strong reducing agent such as Zn), the reduction
products of HNO 3 may actually be N 2 or NH 3 .
Likewise, it is a common error to think that a sulfide, if it reacts with an acid,
will always produce H 2 S. This is not true if the acid is HNO 3 ; the gaseous
products will be NO or NO 2 (depending on the concentration), and sulfur will
be formed. The NO 3 half-reaction is below the H 2 S (or S
2 ~) half-reaction.
Concentrated HNO 3 is so powerful an oxidizing agent that almost all the really
difficult soluble sulfides can be dissolved through oxidation, even though the
same sulfides remain untouched by those strong acids that would lead to the
formation of H 2 S.
It is worth noting that a large number of reducing agents (most of those in the
first eight groups of the abbreviated electron-transfer table!) are unstable in
solution if not protected from the air. Even in neutral water solution, the effect
of air oxidation may be very marked, because O 2 is a strong oxidizing agent. In
some cases the effects may be complicated because of low O 2 pressure or
unusual hydration effects.
By now it may have become a matter of some concern to you that aqueous
solutions of such strong oxidizing agents as KMnO 4 and Na 2 Cr 2 O 7 are stable for
indefinite periods of time. In fact, it would appear that no oxidizing agent that
lies below the half-reaction
4e~ + 4H
+ + O 2 ?± 2H 2 O
E° = 1.23 volts
could exist in water without decomposing the water to liberate O 2 gas. Because
water is not decomposed by these substances, it is evident that, although the
equilibrium position of such reactions does lie far in the direction favoring the
evolution of O 2 , the activation energy must be extremely high and the rate must
be vanishingly small. The detailed reasons why this should be so are still
unknown, despite enormous research effort.
Electrochemistry II: Balancing Equations
are commonly used in the laboratory because they are so soluble and readily
available, and because they are not appreciably oxidizing in neutral aqueous
solution.
We have already seen that strongly acid solutions will make the oxygencontaining negative ions even stronger oxidizing agents; the nitrate ion in an
acid solution is no exception. When concentrated HNO 3 is used, the half-cell
potential is comparable to that of MnOj, but the gaseous product is no longer
NO; it is NO 2 . For concentrated HNO 3 , the half-reaction that must be used is
e- + 2H+ + NOJ ** H 2 O + NO 2
It is a common error to think that a metal, if it reacts with an acid, will produce
H 2 gas. This is not true if the acid is HNO 3 ; the gaseous products will be NO or
NO 2 , depending on the concentration. Under certain circumstances (usually
involving dilute solution and a strong reducing agent such as Zn), the reduction
products of HNO 3 may actually be N 2 or NH 3 .
Likewise, it is a common error to think that a sulfide, if it reacts with an acid,
will always produce H 2 S. This is not true if the acid is HNO 3 ; the gaseous
products will be NO or NO 2 (depending on the concentration), and sulfur will
be formed. The NO 3 half-reaction is below the H 2 S (or S
2 ~) half-reaction.
Concentrated HNO 3 is so powerful an oxidizing agent that almost all the really
difficult soluble sulfides can be dissolved through oxidation, even though the
same sulfides remain untouched by those strong acids that would lead to the
formation of H 2 S.
It is worth noting that a large number of reducing agents (most of those in the
first eight groups of the abbreviated electron-transfer table!) are unstable in
solution if not protected from the air. Even in neutral water solution, the effect
of air oxidation may be very marked, because O 2 is a strong oxidizing agent. In
some cases the effects may be complicated because of low O 2 pressure or
unusual hydration effects.
By now it may have become a matter of some concern to you that aqueous
solutions of such strong oxidizing agents as KMnO 4 and Na 2 Cr 2 O 7 are stable for
indefinite periods of time. In fact, it would appear that no oxidizing agent that
lies below the half-reaction
4e~ + 4H
+ + O 2 ?± 2H 2 O
E° = 1.23 volts
could exist in water without decomposing the water to liberate O 2 gas. Because
water is not decomposed by these substances, it is evident that, although the
equilibrium position of such reactions does lie far in the direction favoring the
evolution of O 2 , the activation energy must be extremely high and the rate must
be vanishingly small. The detailed reasons why this should be so are still
unknown, despite enormous research effort.