Question

Which of the following species will be oxidized by \(1 \mathrm{M} \mathrm{HBr} ?\) (a) \(\mathrm{Na}\) (b) \(\mathrm{Hg}\) (c) \(\mathrm{Pb}\) (d) \(\mathrm{Mn}^{2+}\)

Short answer

Answer: Species (a) Sodium (Na), (b) Mercury (Hg), and (c) Lead (Pb) will be oxidized by a 1 M HBr solution.

Step-by-step solution

Look up standard reduction potentials

Check the Standard Reduction Potentials table for the species involved. The half-reactions we are looking for are: (a) \(\mathrm{Na}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Na}\) with \(E^\circ (\mathrm{Na}^{+}/\mathrm{Na}) = -2.71 \mathrm{V}\) (b) \(\mathrm{Hg}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Hg}\) with \(E^\circ (\mathrm{Hg}^{2+}/\mathrm{Hg}) = 0.85 \mathrm{V}\) (c) \(\mathrm{Pb}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Pb}\) with \(E^\circ (\mathrm{Pb}^{2+}/\mathrm{Pb}) = -0.13 \mathrm{V}\) (d) \(\mathrm{Mn}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}\) with \(E^\circ (\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}) = 1.57 \mathrm{V}\)

Identify the reduction potential for Br-

We should identify the reduction potential for Br- as well: \(\mathrm{Br}_2 + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}\) with \(E^\circ (\mathrm{Br}_2/\mathrm{Br}^-) = 1.07 \mathrm{V}\)

Compare Reduction Potentials

In order for a redox reaction to occur, the reduction potential of the species being reduced must be greater than the reduction potential of the oxidizing species. Comparing the 4 given species: (a) \(E^\circ (\mathrm{Na}^{+}/\mathrm{Na}) = -2.71 \mathrm{V} < 1.07 \mathrm{V}\) (Reaction can occur) (b) \(E^\circ (\mathrm{Hg}^{2+}/\mathrm{Hg}) = 0.85 \mathrm{V} < 1.07 \mathrm{V}\) (Reaction can occur) (c) \(E^\circ (\mathrm{Pb}^{2+}/\mathrm{Pb}) = -0.13 \mathrm{V} < 1.07 \mathrm{V}\) (Reaction can occur) (d) \(E^\circ (\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}) = 1.57 \mathrm{V} > 1.07 \mathrm{V}\) (Reaction cannot occur)

Determine the Species That Will Be Oxidized

From the previous step, we can conclude that species (a) Sodium (Na), (b) Mercury (Hg), and (c) Lead (Pb) will be oxidized by 1 M HBr solution, while (d) \(\mathrm{Mn}^{2+}\) will not be oxidized.