# Redox Titrations

## Redox Titrations

As transition metals are able to form compounds with variable oxidation states, they are able to take part in many redox reactions. Redox reactions involve the transfer of electrons from one species to another.

### Redox Behaviour in Iron

In the presence of air, $Fe^{2+}$ is readily oxidised to the more stable $Fe^{3+}$.

\eqalign{ &Fe^{2+}_{(aq)} \rightleftharpoons &Fe^{3+}_{(aq)} + e^{-} \\ &\text{green} &\text{yellow} }

• $Fe^{3+}$ is an oxidising agent as it can oxidise other compounds by gaining an electron. In turn, $Fe^{3+}_{(aq)}$ is reduced to $Fe^{2+}_{(aq)}$.
• $Fe^{2+}$ is a reducing agent as it can reduce other compounds by donating an electron. In turn, $Fe^{2+}_{(aq)}$ is oxidised to $Fe^{3+}_{(aq)}$.

An example of a redox equilibrium concerning iron(II) is shown below:

$$Fe^{3+}_{(aq)} + 2I^{-}_{(aq)} \rightleftharpoons Fe^{2+}_{(aq)} + I_{2(aq)}$$

### Redox Behaviour in Manganese

In an acid solution, manganese is reduced from an oxidation number of $+7$ in the manganate(VII) ion to $+2$ in the manganese(II) ion.

\eqalign{ &MnO_{4(aq)}^{-} + 8H^{+}_{(aq)} + 5e^{-} \rightleftharpoons &Mn^{2+}_{(aq)} + 4H_{2}O_{(l)} \\ &\text{purple} &\text{pale pink} }

• $MnO_{4}^{-}$ is an oxidising agent as it can oxidise other compounds by gaining an electron. It is commonly used to oxidise solutions containing iron(II).
• $Mn^{2+}_{(aq)}$ is a reducing agent as it can reduce other compounds by donating an electron.

### Redox Behaviour in Chromium

The redox behaviour of chromium is commonly used in organic chemistry to indicate that an alcohol or aldehyde has been oxidised.

\eqalign{ &Cr_{2}O_{7(aq)}^{2-} + 14H^{+}_{(aq)} + 6e^{-} \rightarrow &2Cr^{3+}_{(aq)} + 7H_{2}O_{(l)} \\ &\text{orange} &\text{green} }

## Titrations Involving $MnO_{4}^{-}$ and $Fe^{2+}$

$MnO_{4}^{-}$ is typically used to oxidise solutions containing iron(II) ions. The solution is acidified with sulphuric acid. Hydrochloric acid cannot be used as it reacts with $MnO_{4}^{-}$ ions forming chlorine gas. The reaction for this titration is shown below:

$$MnO^{-}_{4(aq)} + 5Fe^{2+}_{(aq)} + 8H^{+}_{(aq)} \rightarrow Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 4H_{2}O_{(l)}$$

The purple $MnO^{-}_{4(aq)}$ is added to the $Fe^{2+}_{(aq)}$ in the flask. The end point is when there is excess $MnO^{-}_{4(aq)}$ and a permanent pink colour forms.

## Worked Example

A student analyses a sample of sodium sulfite ($Na_{2}SO_{3}$) using the following method.

1. The student dissolves $0.720~g$ of impure sodium sulfite in water.
2. The solution is made up to $100.0~cm^{3}$.
3. The student titrates $25.0~cm^{3}$ of this solution with $0.0200~mol~dm^{-3}$ $KMnO_{4(aq)}$ under acidic conditions. The volume of $KMnO_{4(aq)}$ required to reach the end point is $26.2~cm^{3}$.

The equation for the reaction is shown below:
$$2MnO_{4}^{-} + 6H^{+} + 5SO_{3}^{~~2-} \rightarrow 2Mn^{2+} + 5SO_{4}^{~~2-} + 3H_{2}O$$

Determine the percentage purity of the sample of sodium sulfite.

1. The first step is to calculate the moles of $KMnO_{4(aq)}$ consumed in the titration.
\eqalign{\text{moles of } KMnO_{4(aq)} &= (26.2 \times 10^{-3})(0.02) \\ &= 5.24 \times 10^{-4}~mol }

2. Using the molar ratios in the equation shown above, the moles of $SO_{3}^{~~2-}$ that were used in the titration can be calculated:
\eqalign{\text{moles of } SO_{3}^{~~2-} &= \frac{5.24 \times 10^{-4}~mol}{2} \times 5 \\ &= 1.31 \times 10^{-3}~mol }

3. The number of moles must be scaled up as a result of taking a $25.0~cm^{3}$ sample from the $100.0~cm^{3}$ solution.
\eqalign{\text{moles of } SO_{3}^{~~2-} &= 1.31 \times 10^{-3} \times 4 \\ &= 5.24 \times 10^{-3}~mol}

4. The mass of $Na_{2}SO_{3}$ in the original sample can now be calculated:
\eqalign{ \text{Mr of } Na_{2}SO_{3} &= (23 \times 2) + 32.1 + (16 \times 3) \\ &= 126.1 } \\ \eqalign{ \text{mass of } Na_{2}SO_{3} &= (126.1)(5.24 \times 10^{-3}~mol) \\ &= 0.661g}

5. The percentage purity of the sample can now be calculated:
\eqalign{ \% \text{ purity} &= \frac{0.661}{0.720} \times 100 \\ &= 91.8\% }

## Thiosulfate Titrations

Sodium thiosulfate, containing $S_{2}O_{3}^{~~2-}$ ions, can be titrated against iodine. The reaction is shown below:
$$2S_{2}O_{3(aq)}^{~~2-} + I_{2(aq)} \rightarrow 2I^{-}_{(aq)} + S_{4}O_{6(aq)}^{~~2-}$$

The concentration of iodine in a solution can be determined by titrating it against sodium thiosulfate of known concentration. The number of moles of an oxidising agent that reacts with the produced $I^{-}_{(aq)}$ ions can then be determined.

### Copper Content of Solutions and Alloys

The above method can be used to estimate the concentration of $Cu^{2+}$ ions in a solution or an alloy. The procedure is shown below:

1. First the solution containing $Cu^{2+}$ ions is mixed with iodide ions ($I^{-}_{(aq)}$). The copper(II) ions ($Cu^{2+}$) are reduced to copper ($Cu$) while the iodide ions ($I^{-}$) ions are oxidised to iodine ions ($I_{2}$):
$$2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \rightarrow 2CuI_{(s)} + I_{2(aq)}$$

2. The mixture can then be titrated against a known concentration of sodium thiosulfate. This allows the moles of reacted iodine to be calculated.
$$2S_{2}O_{3(aq)}^{~~2-} + I_{2(aq)} \rightarrow 2I^{-}_{(aq)} + S_{4}O_{6(aq)}^{~~2-}$$
A small amount of starch can be added as an indicator. The solution turns dark blue. The end point is when the dark blue solution sharply turns colourless.

3. By linking the equations together, the unknown number of moles of copper(II) ions can be calculated. This is because:
$$2Cu^{2+} \equiv I_{2} \equiv 2S_{2}O_{3}^{2-}$$

## Worked Example

A sample of bronze was analysed to find the proportion of copper it contained. $0.500~g$ of the bronze was reacted with nitric acid to give a solution containing $Cu^{2+}$ ions. The solution was neutralised and reacted with iodide ions to produce iodine. The iodine was titrated with $22.40~cm^{3}$ of $0.200~mol~dm^{-3}$ sodium thiosulfate.

Calculate the percentage of copper in the bronze.

\eqalign{ Cu_{(s)} &\rightarrow Cu^{2+}_{(aq)} + 2e^{-} \\ 2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} &\rightarrow 2CuI_{(s)} + I_{2(aq)} \\ 2S_{2}O_{3(aq)}^{~~2-} + I_{2(aq)} &\rightarrow 2I^{-}_{(aq)} + S_{4}O_{6(aq)}^{~~2-} }

1. First the number of moles of sodium thiosulfate can be calculated:
\eqalign{\text{moles of } S_{2}O_{3(aq)}^{~~2-} &= (22.4 \times 10^{-3})(0.02) \\ &= 4.48 \times 10^{-3}~mol }
2. By comparing the ratios across the three equations it can be found that:

• $2$ moles of $S_{2}O_{3(aq)}^{~~2-}$ reacted with $1$ mol of $I_{2(aq)}$.
• $1$ mol of $I_{2(aq)}$ was formed from $2$ moles of $Cu^{2+}_{(aq)}$.
• $1$ mol of $Cu^{2+}_{(aq)}$ was formed from $1$ mol of $Cu$.
Therefore, as the ratio is symmetric ($2:1:2$), $1$ mol of $Cu_{(s)}$ is equivalent to $1$ mol of $S_{2}O_{3(aq)}^{~~2-}$.
$$\text{moles of } Cu_{(s)} = 4.48 \times 10^{-3}~mol$$
3. Now the mass of the copper in the sample can be calculated.
\eqalign{ \text{Mr of } Cu_{(s)} &= 63.5 \\ \text{mass of } Cu_{(s)} &= (4.48 \times 10^{-3})(63.5) \\ &= 0.284~g }

4. Finally, the percentage of copper in the bronze can be calculated:
\eqalign{ \% \text{ copper in bronze} &= \frac{0.284}{0.500} \times 100 \\ &= 56.9\% }

## Worked Example

Redox titrations using $KMnO_{4}$ in acidic conditions can be used to analyse reducing agents. Acidified $KMnO_{4}$ is a strong oxidising agent, readily removing electrons.

$$MnO_{4}^{-} + 8H^{+}_{(aq)} + 5e^{-} \rightleftharpoons Mn^{2+} + 4H_{2}O_{(l)}$$

A student analysed a solution of hydrogen peroxide ($H_{2}O_{2(aq)}$) using a redox titration with $KMnO_{4}$, under acidic conditions. Under these conditions, $H_{2}O_{2(aq)}$ is a reducing agent. The overall equation for the reaction is given below.

$$2MnO_{4}^{-} + 6H^{+} + 5H_{2}O_{2} \rightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}$$

The student diluted $25.0~cm^{3}$ of a solution of hydrogen peroxide with water and made the solution up to $250.0~cm^{3}$. The student titrated $25.0~cm^{3}$ of this solution with $0.0200 ~mol~dm^{–3}$ $KMnO_{4}$ under acidic conditions. The volume of $KMnO_{4(aq)}$ required to reach the end-point was $23.45~cm^{3}$.

• Calculate the concentration, in $g~dm^{–3}$, of the undiluted hydrogen peroxide solution.
• What volume of oxygen gas, measured at RTP, would be produced during this titration?
1. First the moles of $KMnO_{4(aq)}$ can be calculated using the known concentration and volume from the titration.
\eqalign{\text{moles of } KMnO_{4(aq)} &= (23.45 \times 10^{-3})(0.02) \\ &= 4.69 \times 10^{-4}~mol }

2. Using the molar ratios in the equation shown above, the moles of $H_{2}O_{2}$ that reacted in the titration can be calculated.
\eqalign{\text{moles of } H_{2}O_{2} &= \frac{4.69 \times 10^{-4}}{2} \times 5 \\ &= 1.17 \times 10^{-3}~mol }

3. As a result of taking a $25.0~cm^{3}$ sample from the $250.0~cm^{3}$ solution, the number of moles of $H_{2}O_{2}$ in the $250.0~cm^{3}$ solution must be determined.
\eqalign{\text{moles of } H_{2}O_{2} &= (1.17 \times 10^{-3}) \times 10 \\ &= 1.17 \times 10^{-2}~mol }

4. The concentration of $H_{2}O_{2}$ is required in $g~dm^{–3}$, therefore the mass of $H_{2}O_{2}$ must be determined. This mass can be used to calculate the concentration of the original $25.0~cm^{3}$ undiluted solution.
\eqalign{ \text{Mr of } H_{2}O_{2} &= 34 \\ \text{mass of } H_{2}O_{2} &= (1.17 \times 10^{-2})(34) \\ &= 0.398~g \\ \text{concentration of } H_{2}O_{2} &= \frac{0.398}{25 \times 10^{-3}} \\ &= 15.9~g~dm^{–3} }

5. For the second part of the question, the number of moles of oxygen formed during the titration should be calculated.
\eqalign{\text{moles of } O_{2} &= \frac{4.69 \times 10^{-4}}{2} \times 5 \\ &= 1.17 \times 10^{-3}~mol }

6. The volume of oxygen gas, measured at RTP can be calculated as shown below:
\eqalign{\text{volume of } O_{2} &= (24.0) \times (\text{moles}) \\ &= (24.0) \times (1.17 \times 10^{-3}) \\ &= 2.8 \times 10^{-2} dm^{3}}

Redox titrations are often applied to unfamiliar situations such as determining the oxidation state of a compound. An example is shown below.

## Worked Example

A student carries out an investigation into the oxidation states of vanadium as outlined below.

1. A $0.126~g$ sample of vanadium metal is completely reacted with acid to form a yellow solution. The solution is made up to $50.0 ~cm^{3}$ in a volumetric flask. This yellow solution contains $VO_{3}^{-}$ ions with vanadium in the $+5$ oxidation state.

2. The yellow solution is reduced to form a violet solution containing $V^{n+}$ ions. This $50.0~cm^{3}$ violet solution contains vanadium in the $+n$ oxidation state.

3. $10.0~cm^{3}$ of the violet solution is titrated with $2.25 \times 10^{–2}~mol~dm^{-3}$ $KMnO_{4(aq)}$. $13.2~cm^{3}$ of $KMnO_{4(aq)}$ is required to reach the end-point.

In the titration, $V^{n+}$ ions are oxidised back to $VO_{3}^{-}$ ions. $MnO_{4}^{-}$ ions are reduced:
$$MnO_{4(aq)}^{-} + 8H^{+}_{(aq)} + 5e^{-} \rightarrow Mn^{2+}_{(aq)} + 4H_{2}O_{(l)}$$

Determine the value of $n$ in the $V^{n+}$ ions formed in Stage 2. Construct an equation for the reaction that takes place during the titration.

1. First the moles of $KMnO_{4(aq)}$ can be calculated using the known concentration and volume from the titration.
\eqalign{\text{moles of } KMnO_{4(aq)} &= (2.25 \times 10^{-2})(13.2 \times 10^{-3}) \\ &= 2.97 \times 10^{-4}~mol }

2. The moles of $V^{n+}$ initially used in the $50.0~cm^{3}$ solution can be calculated using the known mass and $Mr$ of vanadium.
\eqalign{\text{moles of } V^{n+} &= \frac{0.126}{50.9} \\ &= 2.48 \times 10^{-3}~mol }

3. As only a fifth of the solution is used for the titration, the number of moles of $V^{n+}$ used in the titration should be calculated:
\eqalign{\text{moles of } V^{n+} &= \frac{2.48 \times 10^{-3}}{5} \\ &= 4.96 \times 10^{-4}~mol }

4. The ratio between the number of moles of $V^{n+}$ and $KMnO_{4(aq)}$ in the titration can be calculated:
\eqalign{ \frac{\text{moles of } V^{n+}}{\text{moles of } MnO^{-}_{4(aq)}} &= \frac{4.96 \times 10^{-4}}{2.97 \times 10^{-4}} \\ &= \frac{5}{3}}

5. This shows that $5$ moles of $V^{n+}$ reacted with $3$ moles of $MnO^{-}_{4(aq)}$. The half equation for $MnO^{-}_{4(aq)}$ can now be shown as:
$$3MnO_{4(aq)}^{-} + 24H^{+}_{(aq)} + 15e^{-} \rightarrow 3Mn^{2+}_{(aq)} + 12H_{2}O_{(l)}$$
This shows that $15$ electrons must be involved in the $V^{n+}$ half equation. The half equation can be constructed by balancing the number of hydrogen and oxygen atoms.
\eqalign{ 5V^{n+}_{(aq)} &\rightarrow 5VO_{3(aq)}^{-} + 15e^{-} \\ 5V^{n+}_{(aq)} + 15H_{2}O_{(l)} &\rightarrow 5VO_{3(aq)}^{-} + 30H^{+} + 15e^{-}}
The charges can now be balanced:
\eqalign{ 5n &= (-5) + 30 + (-15) \\ n &= 2 }

6. The two half equations can be shown as:
\eqalign{ 3MnO_{4(aq)}^{-} + 24H^{+}_{(aq)} + 15e^{-} &\rightarrow 3Mn^{2+}_{(aq)} + 12H_{2}O_{(l)} \\ 5V^{2+}_{(aq)} + 15H_{2}O_{(l)} &\rightarrow 5VO_{3(aq)}^{-} + 30H^{+} + 15e^{-}}
These two half equations can be combined together to give the overall titration equation.
$$5V^{2+}_{(aq)} + 3MnO_{4(aq)}^{-} + 3H_{2}O_{(l)} \rightarrow 5VO_{3(aq)}^{-} + 3Mn^{2+}_{(aq)} + 6H^{+}$$