Activity 1F (Textbook) – Chemistry Form 5 Chapter 1 (Redox Equilibrium)

Activity 1F:
Carry out a Think-Pair-Share activity.
1. Refer to page 24 for the standard electrode potential cell value, E⁰.

2. Discuss with your friends whether the following reactions occur:
(a) Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Cl⁻(aq) → 2Cr³⁺(aq) + 3Cl₂(g) + 7H₂O(l)

(b) H₂O₂(aq) + 2Br⁻(aq) + 2H⁺(aq) → Br₂(aq) + 2H₂O(l)

3. Present the results of your discussion in class.

Answer:
 2.(a)
– The standard electrode potential value, E⁰ is arranged from the most negative to the most positive.
$$\begin{array}{ll} \mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) & \mathrm{E}^0=+1.33 \mathrm{~V} \\ \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Cl}^{-}(\mathrm{aq}) & \mathrm{E}^0=+1.36 \mathrm{~V} \end{array}$$
– The E₀ value of Cr₂O7²⁻ is less positive. Therefore, Cr₂O7²⁻ ions on the left side are weaker oxidising agents. Cr₂O7²⁻ ions are difficult to gain electrons and reduction reaction does not occur.

– E⁰ value of Cl⁻ is more positive. Therefore, Cl⁻ ions on the right side are weaker reducing agents. Cl⁻ ions are difficult to lose electrons and oxidation reaction does not occur. Therefore, the reaction between Cr₂O7²⁻ and Cl⁻ does not occur.

(b)
$$\begin{aligned} &\text { – The standard electrode potential value, } \mathrm{E}^0 \text { is arranged from the most negative to the most positive. }\\ &\begin{array}{ll} \mathrm{Br}_2(\mathrm{l})+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Br}^{-}(\mathrm{aq}) & \mathrm{E}^0=+1.07 \\ \mathrm{H}_2 \mathrm{O}_2(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) & \mathrm{E}^0=+1.77 \end{array} \end{aligned}$$
– The E⁰ value of H₂O₂ is more positive. Therefore, H₂O₂ on the left side is a stronger oxidising agent. H₂O₂ gains electrons easily and reduction reaction occurs.

– The E⁰ value of Br⁻ is less positive. Therefore, Br⁻ ions on the right side are stronger reducing agents. Br⁻ ions lose electrons easily and oxidation reaction occurs. Therefore, the reaction between H₂O₂ and Br⁻ occurs.