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.
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.