Redox reactions

Oxidising agents and reducing agents

Oxidising agents (or oxidants) can become reduced (gains electrons) while causing another reactant to become oxidised (lose electrons). Reducing agents (or reductants) reduce another reactant by offering electrons to the other reactant. Thus, the reducing agent becomes oxidised (by losing electrons) while reducing another reactant (causing it to gain electrons).

Oxidation number

The oxidation number is the apparent electrical charge on the atom. It helps you keep track of the electrons during a reaction.

• When a chemical species is oxidised, its oxidation number increases.
• When a chemical species is reduced, its oxidation number decreases.

You can assign an oxidation number to any element in a given compound using the rules in the table.

Rule	Example
1. The oxidation number of an atom as an element is zero.	The \(\ce{O}\) in \(\ce{O2}\) has an oxidation number of \(0\). The oxidation numbers of \(\ce{N}\) and \(\ce{Br}\) are \(0\) in \(\ce{N2}\) and \(\ce{Br2}\), respectively. The oxidation number of \(\ce{Fe}\) in \(\ce{Fe}\) metal is also \(0\).
2. The oxidation number of \(\ce{O}\) is usually \(-2\), except when it is bonded to \(\ce{F}\) or in a peroxide (like \(\ce{H2O2}\)).	The \(\ce{O}\) in \(\ce{H2O}\) has an oxidation number of \(-2\). In \(\ce{OF2}\), \(\ce{O}\) has an oxidation number of \(+2\). In \(\ce{H2O2}\), it is \(-1\).
3. The oxidation number of a monoatomic ion is equal to the ion's charge.	The oxidation number of \(\ce{Na}^{+}\) is \(+1\). The oxidation number of \(\ce{Cl}^{-}\) is \(-1\). The oxidation number of \(\ce{S}^{2-}\) is \(-2\).
4. The algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero.	\(\ce{NaCl}\) has an oxidation number of \(0\). The oxidation numbers of \(\ce{Na}\) and \(\ce{Cl}\) are \(+1\) and \(-1\), respectively. \(+1-1=0\)
5. In a polyatomic ion, the sum of the oxidation numbers is equal to the ion's overall charge.	\(\ce{CO}_{3}^{2-}\) has an overall charge of \(-2\). The oxidation numbers of its constituent atoms are \((+4)+(-2)+(-2)+(-2)=-2\).
6. In a compound containing multiple elements, the more electronegative element has its characteristic negative oxidation number and the less electronegative element has a positive oxidation number.	\(\ce{N}\) has an oxidation number of \(-3\), \(\ce{O}\) has an oxidation number of \(-2\) and \(\ce{Cl}\) has an oxidation number of \(-1\). \(\ce{H}\), \(\ce{Ca}\) and \(\ce{Al}\) have the oxidation numbers of \(+1\), \(+2\) and \(+3\), respectively.
7. The oxidation number of \(\ce{H}\) is usually \(+1\) because it is less electronegative than that of the main group elements to which it is normally bonded. With more electropositive elements, \(\ce{H}\) has an oxidation number of \(-1\).	In \(\ce{HCl}\), \(\ce{H}\) has an oxidation number of \(+1\). However, in \(\ce{NaH}\), it has an oxidation number of \(-1\).

Standard reduction potential

Redox reactions involve the movement of electrons, so there is also a voltage associated with it. This is called the standard reduction potential, \(E^{0}\). It is the tendency for a reactant to attract electrons, cause oxidation, and be reduced itself. It is measured in volts (\(\textrm{V}\)) and is defined for an oxidising agent at a concentration of \(1\textrm{M}\) for solutions, \(1\textrm{ atm}\) pressure for gases, and \(25ºC\) temperature.

The greater (more positive) the value of \(E^{0}\), the greater the tendency for the oxidising agent to spontaneously become reduced itself and cause another reactant to undergo oxidation. It will remove electrons from a redox couple with a less positive \(E^{0}\). Thus, there would be a spontaneous redox reaction.

Example 1 – determining oxidation numbers

Find the oxidation number of \(\ce{S}\) in \(\ce{H2SO4}(\textrm{aq})\).

Step 1: Identify the relevant oxidation number rule/s.

Rule 4 states that the algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero, so:

\[(2\times\ce{H})+\ce{S}+(4\times\ce{O})=0\]

Rule \(2\) states that the oxidation number of \(\ce{O}\) is \(-2\) and rule \(6\) states the oxidation number of \(\ce{H}\) is \(+1\).

Step 2: Use algebra to calculate the missing oxidation number.

Let the oxidation number of \(\ce{S}=x\).

\[\begin{align*} (2\times\left(+1\right))+x+(4\times\left(-2\right)) & =0\\
2+x-8 & =0\\
x-6 & =0\\
x & =+6
\end{align*}\]

Therefore, the oxidation number of \(\ce{S}\) in \(\ce{H2SO4}\) is \(+6\).

Example 3 – predicting the products of redox reactions

Predict the products formed from the spontaneous redox reaction between the two redox couples: \(\ce{MnO}_{4}^{-}/\ce{Mn}^{2+}\) and \(\ce{H}_{2}\ce{SO}_{3}/\ce{SO}_{4}^{2-}\), given the data in the table.

Redox couple	\(E^{0}\) (\(\textrm{V}\))	\(e^{-}\) donor/acceptor?	Oxidising or reducing agent?
\(\ce{MnO}_{4}^{-}/\ce{Mn}^{2+}\)	\(+1.51\)	\(e^{-}\) acceptor	Oxidising agent
\(\ce{SO}_{4}^{2-}/\ce{H}_{2}\ce{SO}_{3}\)	\(+0.17\)	\(e^{-}\) donor	Reducing agent

Step 1: Identify the redox couple with the greater/more positive \(E^{0}\).

\(\ce{MnO}_{4}^{-}/\ce{Mn}^{2+}\) has a \(E^{0}\) of \(+1.51\textrm{ V}\).

Step 2: Identify the oxidising agent in this couple. The oxidising agent becomes reduced itself, so its oxidation number decreases.

We can use the worked example above to find the oxidation numbers of each atom in the redox pair.

In \(\ce{MnO}_{4}^{-}\), the oxidation number of \(\ce{Mn}\) is \(+7\) and the oxidation number of \(\ce{O}\) is \(-2\). In \(\ce{Mn}^{2+}\), the oxidation number of \(\ce{Mn}^{2+}\) is \(+2\).

The oxidising agent becomes reduced, so it is the species that contains the atom with the higher oxidation number. Here, it would be \(\ce{MnO}_{4}^{-}\). It has the greater affinity (liking) for electrons. Therefore, \(\ce{MnO}_{4}^{-}\) will be the electron acceptor (i.e. oxidising agent).

Step 3: Identify the reducing agent in the other couple. The reducing agent becomes oxidised itself, so its oxidation number increases.

The sulfur system must donate electrons; hence, \(\ce{H}_{2}\ce{SO}_{3}\) will be the electron donor (i.e. reducing agent). We can check the oxidation numbers to confirm this.

In \(\ce{SO}_{4}^{2-}\), the oxidation numbers of \(\ce{S}\) and \(\ce{O}\) are \(+6\) and \(-2\), respectively. In \(\ce{H2SO3}\), the oxidation numbers of \(\ce{H}\), \(\ce{S}\) and \(\ce{O}\) are \(+1\), \(+4\), and \(-2\), respectively.

The reducing agent would contain the atom with the lower oxidation number, i.e. \(\ce{H2SO3}\).

Step 4: Write the half-equations for the redox reaction.

\(\ce{MnO}_{4}^{-}\) is reduced to \(\ce{Mn}^{2+}\), so the reduction half-equation is:

\[\ce{MnO}_{4}^{-}\rightarrow\ce{Mn}^{2+}\]

\(\ce{H2SO3}\) is oxidised to \(\ce{SO}_{4}^{2-}\), so the oxidation half-equation is:

\[\ce{H}_{2}\ce{SO}_{3}\rightarrow\ce{SO}_{4}^{2-}\]

Example 4 – writing balanced half-equations for redox equations

Balance the following full redox equation for the reaction occurring between the reactants in the previous example:
\[\ce{H}_{2}\ce{SO}_{3}+\ce{MnO}_{4}^{-}\rightarrow\ce{SO}_{4}^{2-}+\ce{Mn}^{2+}\]

Step 1: Assign oxidation numbers to each element in the equation to determine which reactants are being oxidised and reduced.

The oxidation number of \(\ce{Mn}\) in \(\ce{MnO}_{4}^{-}\) is \(+7\) and the \(\ce{Mn}\) oxidation number in \(\ce{Mn}^{2+}\) is \(+2\). Thus, \(\ce{Mn}^{7+}\)is being reduced to \(\ce{Mn}^{2+}\).

The oxidation number of \(\ce{S}\) in \(\ce{H}_{2}\ce{SO}_{3}\) is \(+4\), and the oxidation number of \(\ce{S}\) in \(\ce{SO}_{4}^{2-}\) ion is \(+6\). So, \(\ce{S}\) is being oxidised to \(\ce{SO}_{4}^{2-}\).

Step 2: Divide the equation into the appropriate half-reactions which can be balanced separately.

The reduction half-reaction is: \(\ce{MnO}_{4}^{-}\rightarrow\ce{Mn}^{2+}\)

The oxidation half-reaction is: \[ \ce{H}_{2}\ce{SO}_{3}\rightarrow\ce{SO}_{4}^{2-}\]

Step 3: Write separate balanced chemical equations for reduction and oxidation. Use the acronym KOHES:

• Key elements (elements aside from oxygen and hydrogen)
• Oxygen – add water molecules to balance the oxygen atoms
• Hydrogen – add hydrogen ions (\(\ce{H}^{+}\)) to balance the hydrogen atoms
• Electrons – add electrons (\(e^{-}\)) to balance the charges
• States – add states for all species except electrons, which have no state

Starting with the reduction half-equation, the key elements are already balanced as there is \(1\textrm{ Mg}\) on either side of the reaction arrow. There are \(4\textrm{ O}\) atoms on the reactant side and none on the product side, so to balance the \(\textrm{ O}\) atoms, we need to add \(4\textrm{ H}_{2}\textrm{O}\) to the product side.

\[\ce{MnO}_{4}^{-}\rightarrow\ce{Mn}^{2+}+4\ce{H2O}\]

Next, we need to balance the \(\textrm{H}\) by adding \(8\textrm{ H}^{+}\) ions to the reactant side. This will give us \(8\textrm{ H}\) atoms on each side.

\[\ce{MnO}_{4}^{-}+8\ce{H}^{+}\rightarrow\ce{Mn}^{2+}+4\ce{H2O}\]

To balance the charge, add \(e^{-}\). On the reactant side, the charge is \((-1)+(+8)=(+7)\) and on the product side, the charge is \(+2\). We need to add \(5e^{-}\) to the reactant side to make the charge \(+2\) on both sides.

\[\ce{MnO}_{4}^{-}+8\ce{H}^{+}+5e^{-}\rightarrow\ce{Mn}^{2+}+4\ce{H2O}\]

Finally, we can add states.

\[\ce{MnO}_{4}^{-}(\textrm{aq})+8\ce{H}^{+}(\textrm{aq})+5e^{-}\rightarrow\ce{Mn}^{2+}(\textrm{aq})+4\ce{H2O}(\textrm{l})\]

We can confirm that this is a reduction half-equation by looking at which side of the arrow the electrons are on. For reduction half-equations, the electrons are always on the reactant (left-hand) side. This is indeed the case.

Step 4: Repeat step 3 for the oxidation half-equation.

There is one key element, \(\ce{S}\), and it is already balanced, so we can move on to \(ce{O}\). There are \(3\textrm{ O}\) atoms on the reactant side and \(4\textrm{ O}\) atoms on the product side. We need to add \(1\textrm{ H}_{2}\textrm{O}\) to the reactant side.

\[\ce{H2SO3}+\ce{H2O}\rightarrow\ce{SO}_{4}^{2-}\]

To balance the \(\ce{H}\) atoms, we add \(4\textrm{ H}^{+}\) to the product side.

\[\ce{H2SO3}+\ce{H2O}\rightarrow\ce{SO}_{4}^{2-}+4\ce{H}^{+}\]

To balance the charge, we add \(2e^{-}\) to the product side.

\[\ce{H2SO3}+\ce{H2O}\rightarrow\ce{SO}_{4}^{2-}+4\ce{H}^{+}+2e^{-}\]

And finally, states.

\[\ce{H2SO3}(\textrm{aq})+\ce{H2O}(\textrm{l})\rightarrow\ce{SO}_{4}^{2-}(\textrm{aq})+4\ce{H}^{+}(\textrm{aq})+2e^{-}\]

We can confirm that this is an oxidation question as the electrons are on the product (right-hand) side.

Step 5: To get the full equation, the electrons need to be eliminated. We do this by first making sure that the number of electrons in both equations is equal.

The reduction half-equation has \(5e^{-}\), whereas the oxidation half-equation has \(2e^{-}\). To make them equal, we need to find the least common multiple. In this case, it is 10, so all of the coefficients in the reduction half-equation need to be multiplied by \(2\) and all of the coefficients in the oxidation half-equation need to be multiplied by \(5\).

\[2\times\textrm{( }\ce{MnO}_{4}^{-}(\textrm{aq})+8\ce{H}^{+}(\textrm{aq})+5e^{-}\rightarrow\ce{Mn}^{2+}(\textrm{aq})+4\ce{H2O}(\textrm{l})\textrm{ )}\]
\[5\times\textrm{( }\ce{H2SO3}(\textrm{aq})+\ce{H2O}(\textrm{l})\rightarrow\ce{SO}_{4}^{2-}(\textrm{aq})+4\ce{H}^{+}(\textrm{aq})+2e^{-}\textrm{ )}\]

Adding these together gives:

\[5\ce{H2SO3}(\textrm{aq})+2\ce{MnO}_{4}^{-}(\textrm{aq})+16\ce{H}^{+}(\textrm{aq})+5\ce{H2O}(\textrm{l})+10e^{-}\rightarrow\ce{Mn}^{2+}(\textrm{aq})+5\ce{SO}_{4}^{2-}(\textrm{aq})+20\ce{H}^{+}(\textrm{aq})+8\ce{H2O}(\textrm{aq})+10e^{-}\]

Step 6: Cancel out the equal number of electrons on each side of the equation and simplify the equation by eliminating any chemical species common to each side.

\[5\ce{H2SO3}(\textrm{aq})+2\ce{MnO}_{4}^{-}(\textrm{aq})\rightarrow\ce{Mn}^{2+}(\textrm{aq})+5\ce{SO}_{4}^{2-}(\textrm{aq})+3\ce{H2O}(\textrm{l})+\ce{4H}^{+}(\textrm{aq})\]