Chemistry 9701/41 — May/June 2013

Background Concept

The standard cell potential ($E^\ominus_{\text{cell}}$) is calculated using the standard electrode potentials of the two half-cells involved. The formula is $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$, where the cathode is where reduction occurs (higher/more positive $E^\ominus$) and the anode is where oxidation occurs (lower/less positive $E^\ominus$). Alternatively, you can write the half-equations, reverse the oxidation one and change its sign, then add the potentials.

Understanding the Question

You are given the overall redox equation and asked to calculate $E^\ominus_{\text{cell}}$. You need to identify the two half-reactions, find their standard electrode potentials from the Data Booklet, and apply the correct formula.

Approach

1. Identify the reduction half-reaction: $\text{Fe}^{3+}$ is reduced to $\text{Fe}^{2+}$. $E^\ominus = +0.77 \text{ V}$.
2. Identify the oxidation half-reaction: $\text{I}^-$ is oxidized to $\text{I}_2$. The reduction potential for $\text{I}_2 + 2\text{e}^- \rightarrow 2\text{I}^-$ is $+0.54 \text{ V}$.
3. Calculate $E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 - 0.54$.

Step-by-Step Reasoning

• Look up $E^\ominus$ for $\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}$: $+0.77 \text{ V}$.
• Look up $E^\ominus$ for $\text{I}_2 + 2\text{e}^- \rightarrow 2\text{I}^-$: $+0.54 \text{ V}$.
• Since $\text{Fe}^{3+}$ is reduced and $\text{I}^-$ is oxidized, $E^\ominus_{\text{cell}} = 0.77 - 0.54 = +0.23 \text{ V}$.

Key Takeaways

Always use the formula $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ (reduction minus oxidation) to avoid sign errors. Do not multiply the $E^\ominus$ values by the stoichiometric coefficients.

Common Mistakes

• Multiplying the electrode potential by the stoichiometric coefficient (e.g., using $2 \times 0.77$). Electrode potentials are intensive properties and do not scale with amount.
• Subtracting in the wrong order ($0.54 - 0.77 = -0.23 \text{ V}$), which would incorrectly suggest the reaction is non-spontaneous.

Things to Be Careful About

• Ensure you are using the correct values from the Data Booklet. The value for $\text{Fe}^{3+}/\text{Fe}^{2+}$ is $+0.77 \text{ V}$, not to be confused with $\text{Fe}^{2+}/\text{Fe}$ ($-0.44 \text{ V}$).

Background Concept

The relationship between the standard cell potential and the equilibrium constant $K_c$ is given by $\Delta G^\ominus = -nFE^\ominus_{\text{cell}}$ and $\Delta G^\ominus = -RT \ln K_c$. Combining these gives $E^\ominus_{\text{cell}} = \frac{RT}{nF} \ln K_c$. A positive $E^\ominus_{\text{cell}}$ means $\Delta G^\ominus$ is negative, which means $K_c > 1$. When $K_c > 1$, the equilibrium position lies to the right, meaning products are favoured over reactants.

Understanding the Question

You calculated $E^\ominus_{\text{cell}} = +0.23 \text{ V}$ in part (i). You need to use this value to deduce the position of equilibrium. The word 'hence' means you must base your answer directly on the positive value you just calculated.

Approach

A positive $E^\ominus_{\text{cell}}$ indicates a spontaneous forward reaction under standard conditions and a large $K_c$ value ($K_c > 1$). Therefore, at equilibrium, the concentration of products will be greater than that of reactants.

Step-by-Step Reasoning

• State the reason: $E^\ominus_{\text{cell}}$ is positive ($+0.23 \text{ V}$).
• State the conclusion: More products than reactants (or equilibrium lies to the right / forward reaction is favoured).

Key Takeaways

A positive $E^\ominus_{\text{cell}}$ always correlates with $K_c > 1$ and products being favoured at equilibrium. A negative $E^\ominus_{\text{cell}}$ correlates with $K_c < 1$ and reactants being favoured.

Common Mistakes

• Stating 'the reaction is spontaneous' without linking it to the equilibrium position. The question asks about the equilibrium position, not just spontaneity.
• Forgetting to give a reason. The mark scheme requires both the statement and the reason.

Things to Be Careful About

• Ensure your reason directly references the value from part (i). 'Because $E^\ominus > 0$' is the required phrasing.

Background Concept

The equilibrium constant $K_c$ is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. Pure solids and liquids are excluded, but aqueous ions and dissolved species (like $\text{I}_2(\text{aq})$) are included. The units of $K_c$ are derived by substituting the units of concentration ($\text{mol dm}^{-3}$) into the expression and simplifying.

Understanding the Question

You need to write the $K_c$ expression for the given equilibrium and determine its units. The reaction is:
$2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightleftharpoons 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})$

Approach

1. Write the expression: products over reactants, raised to their coefficients.
2. Substitute units: $([\text{mol dm}^{-3}])^2 \times ([\text{mol dm}^{-3}]) / (([\text{mol dm}^{-3}])^2 \times ([\text{mol dm}^{-3}])^2)$.
3. Simplify the units.

Step-by-Step Reasoning

• Expression (1 mark): Numerator is $[\text{Fe}^{2+}]^2[\text{I}_2]$. Denominator is $[\text{Fe}^{3+}]^2[\text{I}^-]^2$. Ensure the powers match the stoichiometric coefficients (2, 1, 2, 2).
• Units (1 mark):
  Numerator units: $(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3}) = \text{mol}^3 \text{ dm}^{-9}$
  Denominator units: $(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})^2 = \text{mol}^4 \text{ dm}^{-12}$
  $K_c$ units = $\frac{\text{mol}^3 \text{ dm}^{-9}}{\text{mol}^4 \text{ dm}^{-12}} = \text{mol}^{-1} \text{ dm}^3$

Key Takeaways

Always raise concentrations to the power of their stoichiometric coefficients. When calculating units, treat $\text{mol dm}^{-3}$ as a single unit and simplify algebraically.

Common Mistakes

• Forgetting to square the concentrations of $\text{Fe}^{2+}$, $\text{Fe}^{3+}$, and $\text{I}^-$.
• Writing $\text{I}_2$ as a solid or omitting it. It is aqueous ($\text{I}_2(\text{aq})$) in this context.
• Incorrect unit simplification, e.g., writing $\text{mol}^{-1} \text{ dm}^{-3}$ instead of $\text{mol}^{-1} \text{ dm}^3$.

Things to Be Careful About

• The mark scheme allows error carried forward (ecf) from an incorrect expression in (c)(iii) to the units calculation, but you must show the correct derivation for full marks. Ensure your unit calculation matches your expression.

Background Concept

In a closed system where reactants are mixed in stoichiometric proportions (or in this case, equal initial concentrations reacting in a 1:1 ratio), the amounts of products formed and reactants remaining are linked by the stoichiometry of the balanced equation. This allows you to deduce unknown equilibrium concentrations without needing an ICE (Initial, Change, Equilibrium) table if the relationships are clear.

Understanding the Question

You are given that initial solutions of $\text{Fe}^{3+}$ and $\text{I}^-$ had equal concentrations. When mixed in equal volumes ($100 \text{ cm}^3$ each), their concentrations in the mixture are still equal. The reaction consumes them in a 2:2 (or 1:1) ratio. You are given equilibrium concentrations of $[\text{Fe}^{3+}]$ and $[\text{I}_2]$, and need to find $[\text{Fe}^{2+}]$ and $[\text{I}^-]$.

Approach

1. Use the stoichiometry between $\text{Fe}^{2+}$ and $\text{I}_2$: for every 1 $\text{I}_2$ formed, 2 $\text{Fe}^{2+}$ are formed. Since both start at 0, $[\text{Fe}^{2+}] = 2 \times [\text{I}_2]$.
2. Use the stoichiometry between $\text{Fe}^{3+}$ and $\text{I}^-$: they start at equal concentrations and react in a 1:1 ratio. Therefore, the amount remaining must be equal: $[\text{I}^-] = [\text{Fe}^{3+}]$.

Step-by-Step Reasoning

• Calculate $[\text{Fe}^{2+}]$: The reaction produces $\text{Fe}^{2+}$ and $\text{I}_2$ in a 2:1 ratio. Since both were initially 0, at equilibrium $[\text{Fe}^{2+}] = 2 \times [\text{I}_2]$. Given $[\text{I}_2] = 1.0 \times 10^{-2} \text{ mol dm}^{-3}$, $[\text{Fe}^{2+}] = 2 \times 1.0 \times 10^{-2} = 0.02 \text{ mol dm}^{-3}$.
• Calculate $[\text{I}^-]$: The reactants $\text{Fe}^{3+}$ and $\text{I}^-$ start at equal concentrations and are consumed in a 2:2 (1:1) ratio. Therefore, whatever remains must be equal. Given $[\text{Fe}^{3+}] = 2.0 \times 10^{-4} \text{ mol dm}^{-3}$, then $[\text{I}^-] = 2.0 \times 10^{-4} \text{ mol dm}^{-3}$.

Key Takeaways

When reactants are mixed in equal concentrations and react in a 1:1 ratio, their equilibrium concentrations will be equal. When products are formed from zero initial amounts, their equilibrium concentrations are linked by the product stoichiometric ratios.

Common Mistakes

• Trying to calculate the initial concentration and then using an ICE table unnecessarily. While valid, it's more error-prone and slower than using the direct stoichiometric relationships here.
• Forgetting the 2:1 ratio between $\text{Fe}^{2+}$ and $\text{I}_2$.

Things to Be Careful About

• Ensure your units are correct ($\text{mol dm}^{-3}$).
• The mark scheme explicitly states the relationships: '($[\text{Fe}^{2+}]$ must always be twice $[\text{I}_2]$, so) $[\text{Fe}^{2+}] = 0.02$' and '($[\text{I}^-]$ must always be equal to $[\text{Fe}^{3+}]$, so) $[\text{I}^-] = 2 \times 10^{-4}$'.

Background Concept

Once the equilibrium concentrations of all species are known, they can be substituted directly into the $K_c$ expression to calculate the equilibrium constant. This is a straightforward arithmetic calculation, but handling scientific notation and powers correctly is essential to avoid errors.

Understanding the Question

You have the $K_c$ expression from (c)(iii) and all four equilibrium concentrations from (c)(iv) and the question stem. You simply need to substitute and calculate.

Approach

1. Write the $K_c$ expression.
2. Substitute the values: $[\text{Fe}^{2+}] = 0.02$, $[\text{I}_2] = 0.01$, $[\text{Fe}^{3+}] = 2.0 \times 10^{-4}$, $[\text{I}^-] = 2.0 \times 10^{-4}$.
3. Calculate the numerator and denominator separately.
4. Divide and include units.

Step-by-Step Reasoning

• Numerator: $(0.02)^2 \times (1.0 \times 10^{-2}) = (4.0 \times 10^{-4}) \times (1.0 \times 10^{-2}) = 4.0 \times 10^{-6}$
• Denominator: $(2.0 \times 10^{-4})^2 \times (2.0 \times 10^{-4})^2 = (4.0 \times 10^{-8}) \times (4.0 \times 10^{-8}) = 1.6 \times 10^{-15}$
• Division: $\frac{4.0 \times 10^{-6}}{1.6 \times 10^{-15}} = 2.5 \times 10^9$
• Units: From (c)(iii), the units are $\text{mol}^{-1} \text{ dm}^3$.

Key Takeaways

Large $K_c$ values (like $10^9$) confirm the conclusion in part (ii) that products are heavily favoured at equilibrium. Always include units in your final answer for $K_c$ unless they cancel out to give a dimensionless quantity.

Common Mistakes

• Arithmetic errors with scientific notation, especially when squaring terms like $(2.0 \times 10^{-4})^2$.
• Forgetting to include the units in the final answer.
• Rounding errors: ensure you carry full precision through the calculation.

Things to Be Careful About

• The mark scheme allows error carried forward (ecf) from an incorrect expression in (c)(iii) or incorrect concentrations in (c)(iv). If your expression or concentrations were wrong but your substitution and calculation were correct based on those wrong values, you can still earn the method and accuracy marks. However, for the correct answer, use the correct values.
• Significant figures: The data given has 2 significant figures ($2.0$, $1.0$), so the answer should be given to 2 significant figures: $2.5 \times 10^9$.