## Redox reactions - exercises

$\require{mhchem}$

### Redox reactions - practise exercises

#### Determine the oxidation number of each of the following

1. $$\ce{F}$$ in $$\ce{F}_{2}$$

$$0$$

3. $$\ce{Ca}$$ in $$\ce{Ca}^{2+}$$

$$+2$$

5. $$\ce{S}$$ in $$\ce{SO}_{3}$$

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*}

7. $$\ce{P}$$ in $$\ce{PO}_{4}^{3-}$$

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*}

9. $$\ce{Cr}$$ in $$\ce{Cr}_{2}\ce{O}_{3}$$

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*}

11. $$\ce{Al}$$ in $$\ce{AlCl}_{4}^{-}$$

$$+3$$

13. $$\ce{N}$$ in $$\ce{NH}_{4}^{+}$$

$$-3$$

15. $$\ce{Mn}$$ in $$\ce{KMnO}_{4}$$

$$+7$$

17. $$\ce{Fe}$$ in $$\ce{Fe}_{2}\ce{O}_{3}$$

$$+3$$

19. $$\ce{C}$$ in $$\ce{CO}$$

$$+2$$

#### 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. $$\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}+6\ce{O}_{2}\rightarrow6\ce{CO}_{2}+6\ce{H}_{2}\ce{O}$$

1. Species oxidised = $$\ce{C}$$ in $$\ce{CO}$$
Species reduced = $$\ce{Fe}$$ in $$\ce{FeO}$$
Reducing agent - $$\ce{CO}$$
Oxidising agent - $$\ce{FeO}$$

Equation Atom oxidised Atom reduced Oxidising agent Reducing agent
2 $$\ce{Al}$$ $$\ce{Mn}\textrm{ in}\,\ce{MnO}_{2}$$ $$\ce{MnO}_{2}$$ $$\ce{Al}$$
3 $$\ce{Zn}$$ $$\ce{Cu}$$ in $$\ce{CuCl}_{2}$$ $$\ce{CuCl}_{2}$$ $$\ce{Zn}$$
4 $$\ce{I}$$ in $$\ce{KI}$$ $$\ce{Cl}$$ in $$\ce{Cl}_{2}$$ $$\ce{Cl}_{2}$$ $$\ce{KI}$$
5 $$\ce{Mg}$$ $$\ce{C}$$ in $$\ce{CO}_{2}$$ $$\ce{CO}_{2}$$ $$\ce{Mg}$$
6 $$\ce{C}$$ in $$\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}$$ $$\ce{O}$$ in $$\ce{O}_{2}$$ $$\ce{O}_{2}$$ $$\ce{C}_{6}\ce{H}_{12}\ce{O}_{6}$$

#### Balance the following redox equations

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

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.
$6\ce{H}^{+}+2\ce{Al}\rightarrow3\ce{H}_{2}+2\ce{Al}^{3+}$

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

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. $14\ce{H}^{+}+\ce{Cr}_{2}\ce{O}_{7}^{2}+6\ce{Fe}^{2+}\rightarrow2\ce{Cr}^{3+}+7\ce{H}_{2}\ce{O}+6\ce{Fe}^{3+}$

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