Chemistry 9701/43 — October/November 2011

Background Concept

An oxidising agent is a species that accepts electrons. Its oxidising power is measured by its standard electrode potential, $E^\ominus$: the more positive the value, the stronger the oxidising agent. For a half-reaction that involves hydrogen ions, the actual electrode potential depends on $[\text{H}^+]$ as well as on the concentrations of the oxidised and reduced forms.

Understanding the Question

Both $\text{MnO}_4^-$ and $\text{Cr}_2\text{O}_7^{2-}$ are reduced in acidic solution, and their half-equations contain $\text{H}^+$ on the left-hand side. The question asks why increasing the concentration of $\text{H}^+$ increases their oxidising power.

Approach

Write the two half-equations from the Data Booklet. Notice that $\text{H}^+$ is a reactant. Increasing $[\text{H}^+]$ favours the forward (reduction) direction, which makes the electrode potential more positive. A more positive electrode potential corresponds to a stronger oxidising agent.

Step-by-Step Reasoning

• The permanganate half-equation is $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$.
• The dichromate half-equation is $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$.
• In each case $\text{H}^+$ is on the left, so increasing $[\text{H}^+]$ shifts the equilibrium to the right, favouring reduction.
• This makes the electrode potential more positive, so the oxidising power increases.

Key Takeaways

• Acid is not just a spectator in these oxidations; it is a reactant in the half-equation.
• Increasing the concentration of a reactant in a reduction half-reaction makes the reduction more favourable and the electrode potential more positive.

Common Mistakes

• Saying that acid provides $\text{H}^+$ to form water without linking it to the electrode potential.
• Confusing the direction of the shift: increasing $[\text{H}^+]$ favours reduction, not oxidation.
• Forgetting that the oxidising power is directly related to the electrode potential.

Things to Be Careful About

• Use the correct half-equations from the Data Booklet.
• The mark scheme credits any two of: $\text{H}^+$ on the left, $E$ becomes more positive, reaction driven to the right.
• Do not write that increasing $[\text{H}^+]$ makes $E$ more negative.

Background Concept

Transition-metal ions in high oxidation states are often intensely coloured. $\text{MnO}_4^-$ (Mn in +7) is purple/violet; $\text{Cr}_2\text{O}_7^{2-}$ (Cr in +6) is orange. When they act as oxidising agents, they are reduced to lower oxidation states with different colours: $\text{Mn}^{2+}$ is very pale pink/colourless and $\text{Cr}^{3+}$ is green.

Understanding the Question

The question asks for the colour changes observed when each oxidising agent is completely reduced. You need to state the initial colour and the colour after complete reduction.

Approach

Recall the colours of the oxidised and reduced forms. For permanganate, the deep purple disappears to give a colourless (or very pale pink) solution of $\text{Mn}^{2+}$. For dichromate, the orange solution turns green as $\text{Cr}^{3+}$ forms.

Step-by-Step Reasoning

• $\text{KMnO}_4$: the purple/violet $\text{MnO}_4^-$ ion is reduced to $\text{Mn}^{2+}$, which is almost colourless (very pale pink).
• $\text{K}_2\text{Cr}_2\text{O}_7$: the orange $\text{Cr}_2\text{O}_7^{2-}$ ion is reduced to green $\text{Cr}^{3+}$.

Key Takeaways

• Colour changes are diagnostic of the oxidation state of the transition metal.
• In acid, permanganate is reduced to $\text{Mn}^{2+}$; dichromate is reduced to $\text{Cr}^{3+}$.

Common Mistakes

• Saying permanganate turns green (that is dichromate).
• Saying dichromate turns colourless.
• Forgetting that the final permanganate solution may be very pale pink, not completely colourless.

Things to Be Careful About

• The mark scheme allows 'very pale pink' for permanganate.
• Use the exact colours: purple/violet for $\text{MnO}_4^-$, orange for $\text{Cr}_2\text{O}_7^{2-}$, green for $\text{Cr}^{3+}$.

Background Concept

Redox reactions involve transfer of electrons. The oxidation state of an element changes. Manganese in $\text{MnO}_2$ is +4; in $\text{MnSO}_4$ it is +2. Sulfur in $\text{SO}_2$ is +4; in sulfate it is +6. The Data Booklet provides standard electrode potentials that can be used to predict whether a reaction is feasible.

Understanding the Question

This part asks you to use the Data Booklet to suggest an equation for the reaction between $\text{MnO}_2$ and $\text{SO}_2$ in water, and to state the oxidation state changes of manganese and sulfur.

Approach

Identify the half-reactions: $\text{MnO}_2$ is reduced to $\text{Mn}^{2+}$, and $\text{SO}_2$ is oxidised to sulfate. Combine the two half-equations, cancelling electrons and any species that appear on both sides.

Step-by-Step Reasoning

• Reduction: $\text{MnO}_2 + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}$.
• Oxidation: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^-$.
• Adding gives $\text{MnO}_2 + \text{SO}_2 \rightarrow \text{Mn}^{2+} + \text{SO}_4^{2-}$, which is written as $\text{MnSO}_4$.
• Manganese goes from +4 to +2 (reduction); sulfur goes from +4 to +6 (oxidation).

Key Takeaways

• Use the Data Booklet half-equations to construct overall redox equations.
• The oxidation state of an element tells you whether it is oxidised or reduced.

Common Mistakes

• Forgetting to cancel $\text{H}^+$ and $\text{H}_2\text{O}$ when combining half-equations.
• Writing the oxidation state of Mn in $\text{MnO}_2$ as +2.
• Saying sulfur is reduced instead of oxidised.

Things to Be Careful About

• The mark scheme accepts $\text{MnO}_2 + \text{SO}_2 \rightarrow \text{MnSO}_4$ or the ionic form.
• Make sure the equation is balanced in atoms and charge.

Background Concept

For a redox reaction, the position of equilibrium depends on the concentrations of all species that appear in the overall equation. If a species does not appear in the overall equation, changing its concentration has no net effect on the equilibrium position.

Understanding the Question

The pH of the suspension is reduced, meaning $[\text{H}^+]$ is increased. The question asks what effect this has on the extent of the reaction between $\text{MnO}_2$ and $\text{SO}_2$.

Approach

Write the overall equation from part (i) and check whether $\text{H}^+$ appears. If it does not, changing $[\text{H}^+]$ cannot shift the overall equilibrium.

Step-by-Step Reasoning

• The overall equation is $\text{MnO}_2 + \text{SO}_2 \rightarrow \text{MnSO}_4$. No $\text{H}^+$ appears.
• In the reduction half-reaction, $\text{H}^+$ is a reactant; increasing $[\text{H}^+]$ would favour reduction.
• In the oxidation half-reaction, $\text{H}^+$ is a product; increasing $[\text{H}^+]$ would oppose oxidation.
• These two effects cancel exactly, so there is no net effect on the extent of reaction.

Key Takeaways

• Always check the overall equation before predicting the effect of changing a concentration.
• A species can affect individual half-reactions but cancel out in the overall reaction.

Common Mistakes

• Saying that lower pH increases the reaction because acid is needed.
• Forgetting that the oxidation half-reaction produces $\text{H}^+$.
• Not realising that the effects cancel.

Things to Be Careful About

• The mark scheme requires the idea that $\text{H}^+$ is absent from the overall equation or that the two effects cancel.

Background Concept

To construct a redox equation, combine the reduction half-equation and the oxidation half-equation, making sure the number of electrons lost equals the number gained.

Understanding the Question

This part asks for the equation for the reaction between $\text{MnO}_2$ and $\text{Sn}^{2+}$ in acidic solution. You need to use the Data Booklet half-equations.

Approach

Identify the half-reactions: $\text{MnO}_2$ is reduced to $\text{Mn}^{2+}$; $\text{Sn}^{2+}$ is oxidised to $\text{Sn}^{4+}$. Both involve two electrons, so combine directly.

Step-by-Step Reasoning

• Reduction: $\text{MnO}_2 + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O}$.
• Oxidation: $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^-$.
• Adding gives $\text{MnO}_2 + 4\text{H}^+ + \text{Sn}^{2+} \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} + \text{Sn}^{4+}$.

Key Takeaways

• The number of electrons must balance between the two half-equations.
• In acidic solution, $\text{H}^+$ and $\text{H}_2\text{O}$ are used to balance oxygen and hydrogen.

Common Mistakes

• Forgetting to include $\text{H}^+$ in the equation.
• Not balancing the charges.
• Writing $\text{Sn}^{2+}$ as the product instead of $\text{Sn}^{4+}$.

Things to Be Careful About

• The mark scheme gives the balanced equation with $4\text{H}^+$ and $2\text{H}_2\text{O}$.
• Make sure the equation is balanced in atoms and charge.

Background Concept

This is a back-titration. A known amount of $\text{Sn}^{2+}$ is added to the pyrolusite. Some $\text{Sn}^{2+}$ reduces $\text{MnO}_2$; the excess $\text{Sn}^{2+}$ is then titrated with $\text{KMnO}_4$. By finding how much $\text{Sn}^{2+}$ was left over, you can calculate how much reacted with $\text{MnO}_2$, and hence the amount of $\text{MnO}_2$ in the sample.

Understanding the Question

You are given: mass of pyrolusite = 0.100 g; initial moles of $\text{Sn}^{2+}$ = $2.00 \times 10^{-3}$ mol; titration uses $0.0200\text{ mol dm}^{-3}$ $\text{KMnO}_4$ and volume $18.1\text{ cm}^3$. The equation for the titration reaction is given. You need to calculate the percentage of $\text{MnO}_2$.

Approach

1. Calculate moles of $\text{MnO}_4^-$ used in the titration.
2. Use the stoichiometric ratio from the given equation to find moles of $\text{Sn}^{2+}$ that reacted with $\text{MnO}_4^-$.
3. Subtract this from the initial moles of $\text{Sn}^{2+}$ to find moles that reacted with $\text{MnO}_2$.
4. Use the equation from part (i) to convert moles of $\text{Sn}^{2+}$ to moles of $\text{MnO}_2$ (1:1).
5. Convert moles of $\text{MnO}_2$ to mass using $M_r = 86.9$.
6. Divide by the sample mass and multiply by 100 to get the percentage.

Step-by-Step Reasoning

• $n(\text{MnO}_4^-) = c \times V = 0.0200 \times 18.1/1000 = 3.62 \times 10^{-4}\text{ mol}$.
• From $2\text{MnO}_4^- + 5\text{Sn}^{2+} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{Sn}^{4+} + 8\text{H}_2\text{O}$, the ratio is 2:5, so $n(\text{Sn}^{2+}) = 3.62 \times 10^{-4} \times 5/2 = 9.05 \times 10^{-4}\text{ mol}$.
• This is the amount of $\text{Sn}^{2+}$ that reacted with $\text{MnO}_4^-$; the rest reacted with $\text{MnO}_2$: $2.00 \times 10^{-3} - 9.05 \times 10^{-4} = 1.095 \times 10^{-3}\text{ mol}$.
• From part (i), $\text{MnO}_2 + 4\text{H}^+ + \text{Sn}^{2+} \rightarrow \text{Mn}^{2+} + 2\text{H}_2\text{O} + \text{Sn}^{4+}$, so 1 mol $\text{MnO}_2$ reacts with 1 mol $\text{Sn}^{2+}$. Therefore $n(\text{MnO}_2) = 1.095 \times 10^{-3}\text{ mol}$.
• Mass of $\text{MnO}_2 = n \times M_r = 1.095 \times 10^{-3} \times (54.9 + 16 + 16) = 1.095 \times 10^{-3} \times 86.9 = 0.0952\text{ g}$.
• Percentage = $0.0952 / 0.100 \times 100 = 95.2\%$.

Key Takeaways

• In a back-titration, the amount of reagent that reacted with the analyte is the initial amount minus the amount left over.
• Always use the stoichiometric ratio from the balanced equation.
• Pay attention to units: volume in $\text{cm}^3$ must be converted to $\text{dm}^3$.

Common Mistakes

• Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$.
• Using the wrong stoichiometric ratio (e.g. 2:5 instead of 5:2).
• Subtracting the wrong way round.
• Using the molar mass of $\text{MnO}_2$ incorrectly (must be 86.9, not 54.9).

Things to Be Careful About

• The mark scheme gives the percentage as 95%-96%, with 2 or more significant figures.
• Keep track of units and significant figures throughout.
• The final answer should be a percentage with the % sign.