Learning Lab

Redox reactions - practise exercises

Determine the oxidation number of each of the following

1. \(\ce{F}\) in \(\ce{F}_{2}\)

Answer

\(0\)

2. \(\ce{Ca}\) in \(\ce{Ca}^{2+}\)

Answer

\(+2\)

3. \(\ce{S}\) in \(\ce{SO}_{3}\)

Answer

Let’s take the oxidation number of \(\ce{S}\) as \(x\), \[\begin{align*} x+\left(-2\times3\right) & =0\\ x-6 & =0\\ x & =+6 \end{align*}\]

4. \(\ce{P}\) in \(\ce{PO}_{4}^{3-}\)

Answer

Let’s take the oxidation number of \(\ce{P}\) as \(x\) \[\begin{align*} x+\left(-2\times4\right) & =-3\\ x-8 & =-3\\ x & =-3+8\\ x & =+5 \end{align*}\]

5. \(\ce{Cr}\) in \(\ce{Cr}_{2}\ce{O}_{3}\)

Answer

Let’s take an oxidation number of \(\ce{Cr}\) as \(x\) \[\begin{align*} 2x+\left(-2\times3\right) & =0\\ 2x-6 & =0\\ 2x & =6\\ x & =\frac{6}{2}\\ x & =+3 \end{align*}\]

6. \(\ce{Al}\) in \(\ce{AlCl}_{4}^{-}\)

Answer

\(+3\)

7. \(\ce{N}\) in \(\ce{NH}_{4}^{+}\)

Answer

\(-3\)

8. \(\ce{Mn}\) in \(\ce{KMnO}_{4}\)

Answer

\(+7\)

9. \(\ce{Fe}\) in \(\ce{Fe}_{2}\ce{O}_{3}\)

Answer

\(+3\)

10. \(\ce{C}\) in \(\ce{CO}\)

Answer

\(+2\)

For the following redox reactions identify the sing agent, reducing agent, the atom oxidised and the atom reduced.

1. \(\ce{FeO}+\ce{CO}\rightarrow\ce{Fe}+\ce{CO}_{2}\)
2. \(3\ce{MnO}_{2}+4\ce{Al}\rightarrow2\ce{Al}_{2}\ce{O}_{3}+3\ce{Mn}\)
3. \(\ce{Zn}+\ce{CuCl}_{2}\rightarrow\ce{ZnCl}_{2}+\ce{Cu}\)
4. \(\ce{Cl}_{2}+2\ce{KI}\rightarrow2\ce{KCl}+\ce{I}_{2}\)
5. \(2\ce{Mg}+\ce{CO}_{2}\rightarrow2\ce{MgO}+\ce{C}\)
6. \(\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}+6\ce{O}_{2}\rightarrow6\ce{CO}_{2}+6\ce{H}_{2}\ce{O}\)

Balance the following redox equations

1. \(\ce{H}^{+}+\ce{Al}\rightarrow\ce{Al}^{3+}+\ce{H}_{2}\)

Answer

Step 1Divide the equation into the appropriate “half-reactions” which can be balanced separately.
Assign oxidation numbers to each species.

Species	Oxidation number
\(\ce{H}^{+}\)	\(+1\)
\(\ce{Al}\)	\(0\)
\(\ce{Al}^{3+}\)	\(+3\)
\(\ce{H}_{2}\)	\(0\)

Reduction half-reaction: \[ \ce{H}^{+}\rightarrow\ce{H}_{2} \] Oxidation half-reaction: \[ \ce{Al}\rightarrow\ce{Al}^{3+} \]
Step 2- Write a separate balanced chemical equation for each redox-active species. This is the mass balance step.
Reduction half-reaction: \[ 2\ce{H}^{+}\rightarrow\ce{H}_{2} \] Oxidation half-reaction: \[ \ce{Al}\rightarrow\ce{Al}^{3+} \] Step 3 Balance the net charge of each half-reaction by adding electrons.
Reduction half-reaction: \[ 2\ce{H}^{+}+2e\rightarrow\ce{H}_{2} \] Oxidation half-reaction: \[ \ce{Al}\rightarrow\ce{Al}^{3+}+3e \] Step 4 Add two half-reactions in such a way that the electrons cancel out.
Multiply the reduction half-reaction by three and the oxidation half-reaction by two to get an equal number of electrons on both sides.
Reduction half-reaction \[\begin{align*} 3 & \left(2\ce{H}^{+}+2e\rightarrow\ce{H}_{2}\right)\\ 6\ce{H}^{+} & +6e\rightarrow3\ce{H}_{2} \end{align*}\] Oxidation half-reaction: \[\begin{align*} 2 & \left(\ce{Al}\rightarrow\ce{Al}^{3+}+3e\right)\\ 2\ce{Al} & \rightarrow2\ce{Al}^{3+}+6e \end{align*}\] Add two reactions \[\begin{align*} 6\ce{H}^{+}+6e+2\ce{Al} & \rightarrow3\ce{H}_{2}+2\ce{Al}^{3+}+6e \end{align*}\] Step 5 Cancel the electrons and simplify the chemical equation by eliminating any chemical species common to each side of the equation.
\[ 6\ce{H}^{+}+2\ce{Al}\rightarrow3\ce{H}_{2}+2\ce{Al}^{3+} \]

2. \(\ce{Cr}_{2}\ce{O}_{7}^{2-}+\ce{Fe}^{2+}\rightarrow\ce{Cr}^{3+}+\ce{Fe}^{3+}\)

Answer

Step 1 Divide the equation into the appropriate “half-reactions” which can be balanced separately.
Assign oxidation numbers to each species.
The Oxidation number of \(\ce{Cr}\) in \(\ce{Cr}_{2}\ce{O}_{7}^{2-}\):
Let’s take the oxidation number of \(\ce{Cr}\) as \(x\) \[\begin{align*} 2x-14 & =-2\\ 2x & =-2+14\\ 2x & =12\\ x & =+6 \end{align*}\]

Species	Oxidation number
\(\ce{Cr}\) in \(\ce{Cr}_{2}\ce{O}_{7}^{2-}\)	\(+6\)
\(\ce{Fe}^{2+}\)	\(+2\)
\(\ce{Fe}^{3+}\)	\(+3\)
\(\ce{Cr}^{3+}\)	\(+3\)

Reduction half-reaction: \(\ce{Cr}_{2}\ce{O}_{7}^{2-}\rightarrow\ce{Cr}^{3+}\)
Oxidation half-reaction: \(\ce{Fe}^{2+}\rightarrow\ce{Fe}^{3+}\)
Step 2 Write a separate balanced chemical equation for each redox-active species. This is the mass balance step.
Reduction half-reaction: Balance oxygen by adding water to the opposite side of the equation. \[ 14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2-}\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O} \] Oxidation half-reaction: \[ \ce{Fe}^{2+}\rightarrow\ce{Fe}^{3+} \] Step 3Balance the net charge of each half-reaction by adding electrons.
Reduction reaction: \[ 14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2-}+6e\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O} \] Oxidation half-reaction: \[ \ce{Fe}^{2+}\rightarrow\ce{Fe}^{3+}+e \] Step 4 - Add two half-reactions in such a way that the electrons cancel out.
Reduction reaction: \[ 14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2-}+6e\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O} \] Oxidation half-reaction: \[\begin{align*} 6 & \left(\ce{Fe}^{2+}\rightarrow\ce{Fe}^{3+}+e\right)\\ 6\ce{Fe}^{2+} & \rightarrow6\ce{Fe}^{3+}+6e \end{align*}\] Add two half-reactions \[ 14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2}+6e+6\ce{Fe}^{2+}\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O}+6\ce{Fe}^{3+}+6e \] Step 5 - Cancel the electrons and simplify the chemical equation by eliminating any chemical species common to each side of the equation. \[ 14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2}+6\ce{Fe}^{2+}\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O}+6\ce{Fe}^{3+} \]