(a) Draw an energy profile diagram to illustrate a catalysed exothermic reaction and label parts of the curves representing the following: (i) activated com...

Answer Details

(a) Energy profile for a catalysed exothermic reaction

In an exothermic reaction the products lie at a lower energy than the reactants, so the enthalpy change \(\Delta H\) is negative. A catalyst provides an alternative reaction pathway of lower activation energy, so its energy hump (and the activated complex on it) is lower than that of the uncatalysed path. The catalyst changes neither the energy of the reactants nor that of the products, so \(\Delta H\) is exactly the same for both paths.

Reading of the labelled parts:

• (i) Activated complex (without catalyst) - the highest point (top of the upper, uncatalysed curve). It is the unstable transition state through which the reactants pass on the uncatalysed path.
• (ii) Activation energy (with catalyst), \(E_{a(cat)}\) - the smaller energy gap from the reactant level up to the top of the lower (catalysed) curve.
• (iii) Enthalpy change, \(\Delta H\) - the vertical drop from the reactant level down to the product level; it is negative because the reaction is exothermic.

(b) Reasons for the observations

(i) Hydrogen gas is much lighter (less dense) than air. By Graham's law the rate of diffusion of a gas is inversely proportional to the square root of its density, so the light hydrogen molecules diffuse (escape) through the tiny pores of the balloon wall much faster than the heavier oxygen and nitrogen molecules of air. The hydrogen balloon therefore loses its gas and deflates faster.

(ii) Manganese(IV) oxide, \(MnO_2\), acts as a catalyst. It provides an alternative pathway of lower activation energy for the decomposition of hydrogen peroxide, so the reaction that is very slow on its own now proceeds rapidly, giving off oxygen gas quickly as bubbles:

\[2H_2O_2\;\xrightarrow{\;MnO_2\;}\;2H_2O\;+\;O_2\]

The \(MnO_2\) is recovered chemically unchanged at the end.

(iii) Hydrogen chloride is a covalent molecule. In a non-polar solvent such as methylbenzene it stays as neutral \(HCl\) molecules and produces no ions, so there are no free \(H^{+}\) ions and litmus is unaffected. In water it ionises completely, giving hydrogen ions (hydroxonium ions):

\[HCl(g)\;+\;H_2O(l)\;\to\;H_3O^{+}(aq)\;+\;Cl^{-}(aq)\]

These \(H^{+}/H_3O^{+}\) ions make the solution acidic, so it turns blue litmus paper red.

(c) The reaction \(2MnO_4^{-}(aq) + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)\)

(i) Species undergoing reduction: the permanganate ion, \(MnO_4^{-}\). Reason: the oxidation number of manganese decreases from \(+7\) in \(MnO_4^{-}\) to \(+2\) in \(Mn^{2+}\); a decrease in oxidation number is reduction (gain of electrons).

(ii) Reducing agent: the oxalate ion, \(C_2O_4^{2-}\). Reason: it is itself oxidised (the oxidation number of carbon increases from \(+3\) in \(C_2O_4^{2-}\) to \(+4\) in \(CO_2\)), and in being oxidised it supplies the electrons that reduce the \(MnO_4^{-}\).

(iii) Reduction half equation:

\[MnO_4^{-}\;+\;8H^{+}\;+\;5e^{-}\;\to\;Mn^{2+}\;+\;4H_2O\]

(iv) One observation: the purple (pink) colour of the permanganate is discharged, leaving an almost colourless (very pale pink) solution, and effervescence of a colourless gas (\(CO_2\)) is seen.

(d) Electrochemical cell

(i) An electrochemical cell is a device in which a spontaneous chemical (redox) reaction is used to produce electrical energy; that is, it converts chemical energy into electrical energy.

(ii) Three differences between an electrochemical cell and an electrolytic cell: