Chemistry 9701 · AS & A Level · Electrochemistry

Electrochemistry — practice question

Variations in electronegativity can be used to help determine the oxidation number of an atom in different species. The rules that are applied include: - The atom with the higher electronegativity is given a negative oxidation number. - Hydrogen is more electronegative than Group 1 metals. - Oxygen is more electronegative than hydrogen. Which row is correct?

  • A$2\text{CrO}_4^{2-} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$; redox: $\checkmark$; disproportionation: $\times$
  • B$\text{NaH} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$; redox: $\checkmark$; disproportionation: $\checkmark$
  • C$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow \text{MnO}_2 + 2\text{MnO}_4^-$; redox: $\checkmark$; disproportionation: $\checkmark$
  • D$\text{VO}_3^- + 2\text{H}^+ \rightarrow \text{VO}_2^+ + \text{H}_2\text{O}$; redox: $\checkmark$; disproportionation: $\times$

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