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Redox reaction in acidic solution
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Zwei Aufgaben zur Redoxreaktion in saurerer Lösung werden Schritt für Schritt gelöst.
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A complete step by step solution for a redox reaction in acidic solution.Task: Complete the following redox reaction in acidic solution:
}
(M "
(M
32 2
Step 1: Determining oxidation numbers
Since copper is elementary, Cu gets the oxidation number 0:
•
(u
Oxygen usually has the oxidation number -II, except in peroxides or O-F compounds. Since nitric acid has three oxygen atoms
there is a total oxidation state of -VI. Hydrogen usually has the oxidation number +I, except in metal hydrids. Nitric acid has an
overall neutral charge. Therefore, the sum of the oxidation numbers must be zero: N has the oxidation state of +V:
'
+ + _
•
3
} .
Since nitrate ions have a single negative charge, these molecules have an oxidation number of -I. The oxidation number of
copper must be +II to cancel out the two nitrate ions:
•
(u
32
→ 2-
Since nitrogen dioxide is neutrally charged, nitrogen here must be +IV to balance the -IV of the two oxygen atoms.
•
z 2 .
Step 2: Determining the oxidation and reduction reaction
Oxidation occurs when the oxidation number of an atom is increased. In this reaction, that is what happens to the copper.
Since the oxidation number is increased by two, the copper loses two electrons:
Oxidation: (u • (u
32
Ze
This reaction is not yet complete, because you still have to do a mass and charge balance. Since we have two nitrate groups o
the right, we also need two on the left. This also automatically balances the charges completely: there is a total charge of -2 on
the left and the same on the right.
Oxidation: Cu 2 3- • (u
32
2e
Reduction occurs when the oxidation number of an atom decreases. Since the oxidation number of nitrogen decreases by 1,
a reduction takes place here and nitrogen gains one electron in the process:
Reduction: -
e •
3 2
This reaction is also not yet complete. On the left side we have a negative charge. Since we are in an acidic solution we can
use oxonium ions two balance the equation. We need one oxonium ion to balance the electron. Now we still have to balance
the atoms. On the left side there are a total of 4 oxygen atoms and 4 hydrogen atoms. On the right side, however, there are on
2 oxygen atoms and no hydrogen atoms. Since we are in a solution, we can balance the equilibrium with water molecules. On
the right side there are currently 2 oxygen atoms and 4 hydrogen atoms missing, so we need 2 water molecules:
Reduction: -
e • 2-2
3 3 2
In the second last step we still have to balance the electrons. In the oxidation two electrons are given up and
in the reduction only one is taken up. Therefore we have to multiply the reduction reaction with 2.
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