Redox Reactions

Redox reactions have two key characteristics: they involve the transfer of electrons (either gained or lost) and there are changes in oxidation numbers (increasing or decreasing). Either of these can be used to determine whether a half reaction is oxidation or reduction. Half oxidation or reduction equations can be balanced and then combined to form overall redox reactions. Redox chemistry explains the energy conversion processes that drive modern technology, including combustion reactions, and the reactions in galvanic and fuel cells.

Use this page to revise the following concepts within redox reactions:

Determine reducing/oxidising agents

Redox reactions involve simultaneous reduction and oxidation reactions, during which, electron transfer happens and results in changes in oxidation states. If one substance is oxidised, there must also be a substance reduced.

The species undergoing the reduction half-reaction is the oxidising agent (oxidant), and it causes the other species to be oxidised. Similarly, the species undergoing the oxidation half-reaction is the reducing agent (reductant), as it causes the other species to be reduced.

Determine reduction/oxidation reactions by change of oxidation numbers

Determine oxidation numbers

Rule 1: For elements, their oxidation numbers are all 0.

Examples: \(\text{O}\) in \(\text{O}_2: 0\), \(\text{S}\) in \(\text{S}_8: 0\).

Rule 2: The oxidation numbers of monatomic ions are the same as their charges.

Examples: \(\text{Na}^+: +1\), \(\text{O}^{2−}: -2\).

Rule 3: For compounds

  • The more electronegative elements have negative oxidation numbers when combined with other elements.
    Example: oxidation number of \(\text{F}\) in \(\text{ClF}\) is \(-1\).
  • Main group metals have oxidation numbers equal to their ion charges.
    Examples: oxidation number of \(\text{Na}\) in \(\text{NaCl}\) is \(+1\).
  • Hydrogen typically has an oxidation number of \(+1\), but it is \(-1\) in metal hydrides.
    Example: oxidation number of \(\text{H}\) in \(\text{H}_2\text{O}\) is \(+1\), and in \(\text{CaH}_2\) is \(-1\).
  • Oxygen typically has an oxidation number of \(-2\), but it is \(-1\) in peroxides and \(+2\) with fluorine.
    Example: oxidation number of \(\text{O}\) in \(\text{H}_2\text{O}_2\) is \(-1\), and in \(\text{OF}_2\) is \(+2\).

Rule 4: The sum of the oxidation numbers is the same as the charge of the particle. This can be used to solve the oxidation numbers of individual elements.


Worked Example 1

Determining the oxidation number of \(\text{S}\) in \(\text{SO}_2\)

By Rule 3, \(\text{O} = -2\).

By Rule 4, \(\text{SO}_2\) is neutral so the total charge is \(0\).

Let oxidation number of \(\text{S}\) be \(x\), hence \(x + 2(-2) = 0\), thus \(x = + 4\)


Worked Example 2

Determine the oxidation number of \(\text{C}\) in the polyatomic ion \(\ce{CO3^{2-}}\)

By Rule 3, \(\text{O} = -2\).

By Rule 4, the overall charge is \(-2\).

Let the oxidation number of \(\text{C}\) being \(x\).

Hence: \(x + 3 (-2) = -2\), hence \(x = +4\).


Identify oxidising/reduction reaction according to the change in oxidation number

The oxidation number increases in the oxidation half equation, and decreases in the reduction half equation. By analysing the changes in oxidation numbers, we can identify whether a process is oxidation or reduction. This also helps determine whether the reactant is acting as an oxidising agent or a reducing agent.

Worked Example

Determine the oxidising agent and reducing agent in this reaction:

\(\ce{2Fe(s) + 3Cl_2(g) → 2FeCl_3(s)}\)

Solution:

The oxidation number of \(\ce{Fe}\) changes from \(0\) (in \(\ce{Fe}\)) to \(+3\) (in \(\ce{FeCl3}\)). The increase of oxidation number indicates \(\ce{Fe}\) is oxidised, and undergoes oxidation reaction. Therefore, \(\ce{Fe}\) is the reducing agent.

The oxidation number of \(\ce{Cl}\) changes from \(0\) (in \(\ce{Cl2(g)}\)) to \(−1\) (in \(\ce{FeCl3}\)). This decrease in oxidation number indicates that \(\ce{Cl2}\) is reduced, and undergoes a reduction reaction. Therefore, \(\ce{Cl2}\) is the oxidising agent.


Determine reduction/oxidation reactions by electron transfer

The oxidising agent undergoes a reduction reaction, during which it gains electrons from the reducing agent, which undergoes an oxidation reaction.

The following animation demonstrates this transfer of electrons during a redox reaction.

Worked Example

Identify the oxidising agent and reducing agent in the following reaction:

\(\ce{Fe(s) + Cu^{2+}(aq) → Fe^{2+}(aq) + Cu(s)}\)

Solution

  • \(\ce{Fe(s)}\) loses electrons (or increases oxidation number) to form \(\ce{Fe^{2+}(aq)}\), undergoing oxidation, hence \(\ce{Fe(s)}\) is the reducing agent.
  • \(\ce{Cu^{2+}(aq)}\) gains electrons (or decreases oxidation number) to form \(\ce{Cu(s)}\), undergoing reduction, hence \(\ce{Cu^{2+}(aq)}\) is the oxidising agent.

Writing conjugate redox pairs

A conjugate redox pair consists of a reactant and the corresponding product formed through the gain or loss of electrons.

In the reaction:

\(\ce{Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)}\)

  • \(\ce{Fe}\) is oxidised to \(\ce{Fe^{2+}}\), forming the conjugate redox pair \(\ce{Fe^{2+}/Fe}\).
  • \(\ce{Cu^{2+}}\) is reduced to Cu, forming the pair \(\ce{Cu^{2+}/Cu}\).

When writing conjugate redox pairs, the species with the higher oxidation number is often written first, followed by a “/” and then the species with the lower oxidation number, such as \(\ce{Fe^{2+}/Fe}\) and \(\ce{Cu^{2+}/Cu}\).

Balancing redox equations

Half equations show how two substances in a redox pair are converted from one to the other through gaining/losing electrons.

Balancing half equations for simple redox reactions

For this unbalanced equation:

\(\ce{Br2(l) + KI(aq) → KBr(aq) + I2(s)}\)

To write its reduction half equation, follow these steps:

  1. Exclude the spectator ions:
    \(\ce{Br2(l) + I−(aq) → Br−(aq) + I2(s)}\)
  2. Identify reduced chemical species whose oxidation number decreases - it is the oxidising agent: \(\ce{Br2}\)
  3. Identify the product of the oxidising agent: \(\ce{Br2\rightarrow Br^{−}}\)
  4. Balance the number of atoms in the half equation: \(\ce{Br2\rightarrow2 Br^{−}}\)
  5. Add electrons to balance the charge:
    \(\ce{Br2 + 2e^{−}\rightarrow2 Br^{−}}\)
  6. Include states:
    \(\ce{Br2(l) + 2e^{−}\rightarrow 2Br^{−}(aq)}\)
  7. Check: ensure the half-equation is balanced in terms of both atoms and charge.

Similarly, the oxidation half equation is \(\ce{2I−(aq)\rightarrow I2(s) + 2e^{−}}\)

Balancing half equations in acidic aqueous conditions

In an acidic aqueous reaction system, free moving \(\ce{H^{+}(aq)}\) and \(\ce{H2O(l)}\) are available to balance the half equations.

For example: in the reaction between \(\ce{C2H5OH (aq)}\) and \(\ce{Cr2O7^{2-}(aq)}\) in an acidic aqueous solution, forming \(\ce{CH3COOH (aq)}\) and \(\ce{Cr^{3+}(aq)}\), the \(\ce{KOHeS}\) method can be used to write balanced half equations.

For the oxidation half equation,  identify the substance whose oxidation number increases: \(\ce{C2H5OH \rightarrow CH3COOH}\)

  1. K: Balance the key element, which is the element that changes oxidation number.

    Here, the carbon is the key element and it is already balanced as there are two carbon atoms on each side of the arrow.

  2. O: Balance the oxygen atoms by adding \(\ce{H2O}\) to the side that requires more \(O\).

    \(\ce{C2H5OH + H2O  CH3COOH}\)

  3. H: Balance the hydrogen atoms by adding \(\ce{H^{+}}\)to the side that requires more H.

    \(\ce{C2H5OH + H2O  CH3COOH + 4H^{+}}\)

  4. e: Balance the charge by adding electrons (\(\ce{e^{-}}\)) to the more positive side:

    \(\ce{C2H5OH + H2O  CH3COOH + 4H^{+} + 4e^{-}}\)

  5. S: Add correct states:

    \(\ce{C2H5OH(aq) + H2O(l)  CH3COOH(aq) + 4H^{+}(aq) + 4e^{-}}\)

Balance half equations in basic aqueous conditions

To balance complex half-equations in basic conditions, first balance them as if in acidic conditions using the \(\ce{KOHeS}\) method. Then, add \(\ce{OH^{-}}\) to neutralize \(\ce{H^{+}}\), forming \(\ce{H2O}\). Finally, simplify the equation.

Worked Example

\(\ce{MnO2(s)}\) can be converted to \(\ce{Mn2O3(s)}\) in basic aqueous solution. Write the balanced half equation.

Solution

  • Use \(\ce{KOHeS}\) method and balance the half equation in acidic solutions: \(\ce{2MnO2(s) + 2H^{+}(l) + 2e^{-} \rightarrow Mn2O3(s) + H2O (aq)}\)
  • Add \(\ce{OH^{-}}\) to neutralise all \(\ce{H^{+}}\). There are two \(\ce{H^{+}}\) to the left, hence add two \(\ce{OH^{-}}\) to both sides. On the left side, two \(\ce{H2O}\) are formed.

    Hence: \(\ce{2MnO2(s) + 2H2O(l) + 2e^{-} \rightarrow Mn2O3(s) + H2O(l) + 2OH^{-}(aq)}\)
  • Simplify the equation by cancelling the repeating \(\ce{H2O}\) on both sides of the equation: \(\ce{2MnO2(s) + H2O(l) + 2e- \rightarrow Mn2O3(s) + 2OH-(aq)}\)

Balancing overall redox equations

Sometimes, balancing complex overall redox reactions can be challenging. By splitting the overall reaction into half-equations, balancing each half separately, and then recombining them, we can balance the overall redox reactions. When combining half equations, the number of electrons gained or lost from half equations must be the same.  We can balance the number of electrons in each half-equation by multiplying one or both half-reactions.

Worked Example

\(\ce{C_2H_5OH (aq)}\) and \(\ce{Cr_2O_7^{2-}(aq)}\) react in acidic aqueous solutions, and form \(\ce{CH3COOH (aq)}\) and \(\ce{Cr^{3+} (aq)}\).

Step 1: Balance the two half equations:

Reduction: \(\ce{Cr2O7^{2-}(aq) + 14H^{+}(aq) + 6e^{-} 2Cr^{3+}(aq) + 7H2O(l)}\)
Oxidation: \(\ce{C2H5OH(aq) + H2O(l) CH3COOH(aq) + 4H^{+}(aq) + 4e^{-}}\)

Step 2: Balance the electrons in two half cells. This can be done by multiplying all coefficients in the reduction half-equation by \(2\) and in the oxidation half-equation by \(3\).

Reduction\( \times 2\): \(\ce{2Cr2O7^{2-}(aq) + 28H^{+}(aq) + 12e^{-}\rightarrow 4Cr^{3+}(aq) + 14H2O(l)}\)
Oxidation\(\times 3\): \(\ce{3C2H5OH(aq) + 3H2O(l) \rightarrow 3CH3COOH(aq) + 12H^{+}(aq) + 12e^{-}}\)

Step 3: Add two half equations together:

\(\ce{2Cr2O7^{2-}(aq) + 28H^{+}(aq) + 3C_2H_5OH(aq) + 3H_2O(l)\rightarrow}\)

\(\ce{4Cr^{3+}(aq) + 14H_2O(l) + 3CH_3COOH(aq) + 12H^{+}(aq)}\)

Step 4: Simplify by cancelling out particles that appear on both sides:

\(\ce{2Cr_2O_7^{2-}(aq) + 16H^{+}(aq) + 3C2H5OH(aq)\rightarrow 4Cr^{3+}(aq) + 11H_2O(l) + 3CH_3COOH(aq)}\)