Redox Reactions — Diagnostic Tests

Unit Tests

UT-1: Oxidation Number Assignment

Question: Assign oxidation numbers to all elements in: (a) $\text{KMnO}_4$, (b) $\text{K}_2\text{Cr}_2\text{O}_7$, (c) $\text{Na}_2\text{S}_2\text{O}_3$, (d) $\text{Fe}_3\text{O}_4$, (e) $\text{H}_2\text{O}_2$.

Solution:

(a) $\text{KMnO}_4$: K = $+1$ (group 1), O = $-2$ (usually). Let Mn = $x$: $+1 + x + 4(-2) = 0$, $x = +7$. Mn is $+7$.

(b) $\text{K}_2\text{Cr}_2\text{O}_7$: K = $+1$, O = $-2$. $2(+1) + 2x + 7(-2) = 0$, $2 + 2x - 14 = 0$, $2x = 12$, $x = +6$. Cr is $+6$.

(c) $\text{Na}_2\text{S}_2\text{O}_3$: Na = $+1$, O = $-2$. $2(+1) + 2x + 3(-2) = 0$, $2 + 2x - 6 = 0$, $2x = 4$, $x = +2$. Average S oxidation state is $+2$. (In reality, the two S atoms have different oxidation states: the central S is $+6$ and the terminal S is $-2$, averaging to $+2$.)

(d) $\text{Fe}_3\text{O}_4$: O = $-2$. $3x + 4(-2) = 0$, $3x = 8$, $x = +2.67$. Average Fe oxidation state is $+2.67$. $\text{Fe}_3\text{O}_4$ is a mixed oxide containing both $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ in the ratio 1:2 (one $\text{FeO}\cdot\text{Fe}_2\text{O}_3$), giving $(+2 + 2 \times +3)/3 = +2.67$.

(e) $\text{H}_2\text{O}_2$: H = $+1$. $2(+1) + 2x = 0$, $2x = -2$, $x = -1$. O is $-1$ (peroxide exception to the usual $-2$).

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UT-2: Balancing Redox Equations in Acidic Solution

Question: Balance the following redox equation in acidic solution:

$\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$

Solution:

Half-reactions:

Reduction: $\text{MnO}_4^- \to \text{Mn}^{2+}$

Oxidation: $\text{Fe}^{2+} \to \text{Fe}^{3+}$

Balance reduction half-reaction (acidic):

$\text{MnO}_4^- \to \text{Mn}^{2+}$

Balance O with $\text{H}_2\text{O}$: $\text{MnO}_4^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Balance H with $\text{H}^+$: $\text{MnO}_4^- + 8\text{H}^+ \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Balance charge: left = $-1 + 8 = +7$, right = $+2$. Add $5e^-$ to left:

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Balance oxidation half-reaction:

$\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$

Equalise electrons: multiply oxidation by 5:

$5\text{Fe}^{2+} \to 5\text{Fe}^{3+} + 5e^-$

Add half-reactions:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

Check: left charge $= -1 + 8 + 10 = +17$, right charge $= +2 + 15 = +17$. Balanced.

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UT-3: Electrochemical Cell EMF

Question: A voltaic cell is constructed with $\text{Zn}\mid\text{Zn}^{2+}(1.0\ \text{mol dm}^{-3})$ and $\text{Cu}^{2+}(1.0\ \text{mol dm}^{-3})\mid\text{Cu}$. Given $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$ and $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$, calculate the standard cell potential, identify the anode and cathode, write the overall cell equation, and determine the standard Gibbs free energy change.

Solution:

Since $E^\circ(\text{Cu}^{2+}/\text{Cu}) \gt E^\circ(\text{Zn}^{2+}/\text{Zn})$, Cu is reduced (cathode) and Zn is oxidised (anode).

Anode (oxidation): $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ Cathode (reduction): $\text{Cu}^{2+} + 2e^- \to \text{Cu}$

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = +1.10\ \text{V}$

Overall: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)$

$\Delta G^\circ = -nFE^\circ = -(2)(96485)(1.10) = -212267\ \text{J mol}^{-1} = -212\ \text{kJ mol}^{-1}$

The negative $\Delta G^\circ$ confirms the reaction is spontaneous under standard conditions.

Integration Tests

IT-1: Electrochemistry and Equilibrium (with Equilibrium)

Question: For the cell $\text{Zn}\mid\text{Zn}^{2+}\mid\mid\text{Cu}^{2+}\mid\text{Cu}$ with $E^\circ_{\text{cell}} = 1.10\ \text{V}$, calculate the equilibrium constant $K$ for the reaction at $298\ \text{K}$. At what ratio of $[\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ does the cell potential drop to $0.50\ \text{V}$?

Solution:

At equilibrium, $E_{\text{cell}} = 0$, and $\Delta G^\circ = -RT\ln K$.

$nFE^\circ = RT\ln K$

$\ln K = \frac{nFE^\circ}{RT} = \frac{2 \times 96485 \times 1.10}{8.314 \times 298} = \frac{212267}{2477.6} = 85.66$

$K = e^{85.66} = 1.5 \times 10^{37}$

The enormous $K$ confirms the reaction goes essentially to completion.

Using the Nernst equation for $E_{\text{cell}} = 0.50\ \text{V}$:

$E = E^\circ - \frac{0.0592}{n}\log Q$

$0.50 = 1.10 - \frac{0.0592}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$

$\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{(1.10 - 0.50) \times 2}{0.0592} = \frac{1.20}{0.0592} = 20.27$

$\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 10^{20.27} = 1.9 \times 10^{20}$

Even when the cell potential has dropped by more than half, the reaction quotient is astronomically large -- the reaction is still overwhelmingly product-favoured.

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IT-2: Redox Titration and Stoichiometry (with Measurement and Data Processing)

Question: A $25.0\ \text{cm}^3$ sample of hydrogen peroxide solution was acidified and titrated with $0.0200\ \text{mol dm}^{-3}$ $\text{KMnO}_4$. The average titre was $28.45\ \text{cm}^3$. The unbalanced equation is: $\text{MnO}_4^- + \text{H}_2\text{O}_2 + \text{H}^+ \to \text{Mn}^{2+} + \text{O}_2 + \text{H}_2\text{O}$. Calculate the concentration of $\text{H}_2\text{O}_2$ in $\text{g dm}^{-3}$ and determine the volume of oxygen gas (at STP) produced from $100\ \text{cm}^3$ of this solution.

Solution:

Balanced equation: $2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \to 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}$

Moles $\text{MnO}_4^- = 0.0200 \times 0.02845 = 5.690 \times 10^{-4}\ \text{mol}$

From stoichiometry: $n(\text{H}_2\text{O}_2) = \frac{5}{2} \times 5.690 \times 10^{-4} = 1.4225 \times 10^{-3}\ \text{mol}$

Concentration: $c = \frac{1.4225 \times 10^{-3}}{0.0250} = 0.05690\ \text{mol dm}^{-3}$

In $\text{g dm}^{-3}$: $0.05690 \times 34.01 = 1.935\ \text{g dm}^{-3}$

For $100\ \text{cm}^3$: $n(\text{H}_2\text{O}_2) = 0.05690 \times 0.100 = 5.690 \times 10^{-3}\ \text{mol}$

$n(\text{O}_2) = \frac{5}{5} \times 5.690 \times 10^{-3} = 5.690 \times 10^{-3}\ \text{mol}$

$V(\text{O}_2) = nRT/P = \frac{5.690 \times 10^{-3} \times 8.314 \times 273.15}{101325} = \frac{12.93}{101325} = 1.28 \times 10^{-4}\ \text{m}^3 = 0.128\ \text{dm}^3 = 128\ \text{cm}^3$

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IT-3: Corrosion and Electrochemistry (with Periodicity)

Question: Explain why iron corrodes more readily than aluminium, even though aluminium has a more negative standard electrode potential ($E^\circ(\text{Al}^{3+}/\text{Al}) = -1.66\ \text{V}$ vs $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\ \text{V}$). Discuss the role of the oxide layer and periodic trends.

Solution: Thermodynamically, Al should be more reactive than Fe (more negative $E^\circ$). However, kinetics dominate corrosion behaviour.

When aluminium is exposed to air, it rapidly forms a thin (2--10 nm), continuous, adherent layer of $\text{Al}_2\text{O}_3$ on its surface. This oxide layer is impervious to water and oxygen -- it passivates the surface and prevents further oxidation. The layer is self-repairing: if scratched, fresh Al is immediately exposed to air and re-forms the oxide.

Iron also forms an oxide layer ($\text{Fe}_2\text{O}_3/\text{Fe}_3\text{O}_4$), but rust (hydrated iron(III) oxide, $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$) is porous, flaky, and non-adherent. It does not protect the underlying metal; instead, moisture and oxygen can penetrate through cracks and continue the corrosion process. This is why iron corrosion is progressive and destructive.

The periodic trend explanation: Al is in period 3 with a small ionic radius ($\text{Al}^{3+} = 53.5\ \text{pm}$) and high charge density. The $\text{Al}^{3+}$ ion polarises the $\text{O}^{2-}$ ions strongly, forming a tightly bonded, compact oxide lattice. Fe is in period 4 with a larger ionic radius ($\text{Fe}^{2+} = 78\ \text{pm}$, $\text{Fe}^{3+} = 64.5\ \text{pm}$) and the oxide lattice has more defects and is less uniformly bonded, allowing water incorporation and porosity.