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Mobile App Chapter 20: Problem 24
Balance the following redox equations. (a) \(\mathrm{Fe}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{NO}_{2}(g)\) (acidic) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic)
Short Answer
Expert verified
Question: Balance the following two redox equations.
(a) \(Fe(s)+NO_{3}^{-}(aq) \longrightarrow Fe^{3+}(aq)+NO_{2}(g)\) (acidic)
(b) \(Cr(OH)_{3}(s)+O_{2}(g) \longrightarrow CrO_{4}^{2-}(aq)\) (basic)
Answer:
(a) The balanced redox equation in acidic conditions is:
\(Fe(s) + 3NO_{3}^{-}(aq) + 6H^+(aq) \longrightarrow Fe^{3+}(aq) + 3NO_{2}(g) + 3H_2O(l)\)
(b) The balanced redox equation in basic conditions is:
\(2Cr(OH)_{3}(s) + \frac{3}{2}O_{2}(g) \longrightarrow 2CrO_{4}^{2-}(aq) + 12H_2O(l)\)
Step by step solution
01
Identify the oxidation and reduction half-reactions
Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q)\)
Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\)
02
Balance the atoms other than hydrogen and oxygen
The atoms of elements other than hydrogen and oxygen are already balanced.
03
Balance the oxygen atoms by adding water molecules
Oxidation: No need to add water since there is no oxygen atom in this half-reaction.
Reduction: Add 1 water molecule to balance the oxygen atoms.
\(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)
04
Balance the hydrogen atoms by adding hydrogen ions
Oxidation: No need to add hydrogen ions since there is no hydrogen atom in this half-reaction.
Reduction: Add 2 H+ ions to balance the hydrogen atoms.
\(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)
05
Balance the charge of the half-reactions by adding electrons
Oxidation: Add 3 electrons to balance the charge.
\(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3 e^-\)
Reduction: Add 1 electron to balance the charge.
\(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) + e^- \longrightarrow \mathrm{NO}_{2}(g) + \mathrm{H_2O}(l)\)
06
Make the number of electrons in the two half-reactions equal
Multiply the reduction half-reaction by 3.
Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3 e^-\)
Reduction: \(3(\mathrm{NO}_{3}^{-}(a q) + 2\mathrm{H^+}(a q) + e^-) \longrightarrow 3(\mathrm{NO}_{2}(g) + \mathrm{H_2O}(l))\)
So the new reduction half-reaction is:
\(\mathrm{3NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) + 3 e^- \longrightarrow 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\)
07
Add the two half-reactions together
\(\mathrm{Fe}(s) + 3\mathrm{NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\)
08
Cancel out any common terms
There are no common terms that can be canceled out. So the balanced redox equation in acidic conditions is:
\(\mathrm{Fe}(s) + 3\mathrm{NO}_{3}^{-}(a q) + 6\mathrm{H^+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + 3\mathrm{NO}_{2}(g) + 3\mathrm{H_2O}(l)\)
(b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic)
Follow the same steps 1 to 7 as in (a):
09
Identify the oxidation and reduction half-reactions
Oxidation: \(\mathrm{Cr(OH)}_3(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\)
Reduction: \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H_2O}(l)\)
10
Balance the atoms other than hydrogen and oxygen
Oxidation: Add 2 Cr atoms to the product side.
\(2\mathrm{Cr(OH)}_3(s) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q)\)
Reduction: No need to balance atoms since there are no other elements in this half-reaction.
11
Balance the oxygen atoms by adding water molecules
Oxidation: Add 6 water molecules to balance the oxygen atoms.
\(2\mathrm{Cr(OH)}_3(s) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\)
Reduction: Add 4 water molecules to balance the oxygen atoms.
\(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{H_2O}(l)\)
12
Balance the hydrogen atoms by adding hydrogen ions
Oxidation: Add 12 H+ ions to balance the hydrogen atoms.
\(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\)
Reduction: Add 8 H+ ions to balance the hydrogen atoms.
\(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l) + 8\mathrm{H}^+(a q) \longrightarrow \mathrm{H_2O}(l)\)
13
Balance the charge of the half-reactions by adding electrons
Oxidation: Add 6 electrons to balance the charge.
\(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) + 6 e^- \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\)
Reduction: Add 4 electrons to balance the charge.
\(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l)+ 8\mathrm{H}^+(a q) + 4 e^- \longrightarrow \mathrm{H_2O}(l)\)
14
Make the number of electrons in the two half-reactions equal
Multiply the reduction half-reaction by 1.5 (or multiply by 3 and then divide by 2).
Oxidation: \(2\mathrm{Cr(OH)}_3(s) + 12\mathrm{H^+}(a q) + 6 e^- \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{H_2O}(l)\)
Reduction: \((3/2)(\mathrm{O}_{2}(g) + 4\mathrm{H}_2\mathrm{O}(l)+ 8\mathrm{H}^+(a q) + 4 e^-) \longrightarrow (3/2)\mathrm{H_2O}(l)\)
So the new reduction half-reaction is:
\(\frac{3}{2}\mathrm{O}_{2}(g) + 6\mathrm{H}_2\mathrm{O}(l)+ 12\mathrm{H}^+(a q) + 6 e^- \longrightarrow 3\mathrm{H_2O}(l)\)
15
Add the two half-reactions together
\(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) + 6\mathrm{H}_2\mathrm{O}(l)+ 12\mathrm{H}^+(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 9\mathrm{H_2O}(l)\)
Now we need one additional step to account for the basic conditions:
16
Convert hydrogen ions into hydroxide ions
Add 12 OH- ions to both sides of the equation to remove the H+ ions.
\(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) + 12\mathrm{OH}^-(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 9\mathrm{H_2O}(l) + 12\mathrm{OH}^-(a q)\)
Combine the water and hydroxide ions to form 3 more water molecules.
\(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 12\mathrm{H_2O}(l)\)
So the balanced redox equation in basic conditions is:
\(2\mathrm{Cr(OH)}_3(s) + \frac{3}{2}\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 12\mathrm{H_2O}(l)\)
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