9701/13

Chemistry 9701/13May/June 2013

Cambridge AS Level · Multiple Choice (AS Level) · answer key with instant marking and worked solutions

40
questions
40
marks
60
minutes

Topics Atoms, Molecules and Stoichiometry · States of Matter · Hydrocarbons · Introduction to Organic Chemistry · Electrochemistry · Atomic Structure · +12 more

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Q11MElectrochemistryFree sample

In the redox reaction shown, how do the oxidation states of vanadium and sulfur change?

VO2++SO2V3++SO42\text{VO}_2^+ + \text{SO}_2 \rightarrow \text{V}^{3+} + \text{SO}_4^{2-}

Options

vanadium (from)vanadium (to)sulfur (from)sulfur (to)
A+1+30-2
B+1+3+4+6
C+5+30-2
D+5+3+4+6
DifficultyMedium-Easy
Worked solution

Working

Assign oxidation numbers:

  • In VO2+\text{VO}_2^+: each O is -2, total -4, charge +1, so V = +5.
  • In V3+\text{V}^{3+}: V = +3.
  • In SO2\text{SO}_2: each O is -2, total -4, molecule neutral, so S = +4.
  • In SO42\text{SO}_4^{2-}: each O is -2, total -8, charge -2, so S = +6.

Thus V changes from +5 to +3; S changes from +4 to +6.

Answer

D

Final answer

D

Detailed explanation

Background Concept

Oxidation numbers (oxidation states) are a bookkeeping tool to track electron transfer in redox reactions. For a polyatomic ion, the sum of the oxidation numbers of all atoms equals the charge on the ion. Oxygen is usually assigned -2 in compounds (except in peroxides and with fluorine).

Understanding the Question

The reaction is:

VO2++SO2V3++SO42\text{VO}_2^+ + \text{SO}_2 \rightarrow \text{V}^{3+} + \text{SO}_4^{2-}

We need to find the oxidation numbers of vanadium and sulfur on both sides and see how they change.

Approach

  1. Determine the oxidation number of vanadium in VO2+\text{VO}_2^+ and in V3+\text{V}^{3+}.
  2. Determine the oxidation number of sulfur in SO2\text{SO}_2 and in SO42\text{SO}_4^{2-}.
  3. Compare the changes and select the correct option.

Step-by-Step Reasoning

Vanadium in VO2+\text{VO}_2^+:

  • Each oxygen is -2, so two oxygens contribute -4.
  • The ion has a total charge of +1.
  • Let V = x. Then x + (-4) = +1 → x = +5.

Vanadium in V3+\text{V}^{3+}:

  • The ion has a charge of +3, so V = +3.

Sulfur in SO2\text{SO}_2:

  • Each oxygen is -2, so two oxygens contribute -4.
  • The molecule is neutral, so S = +4.

Sulfur in SO42\text{SO}_4^{2-}:

  • Four oxygens contribute -8.
  • The ion has a charge of -2.
  • Let S = y. Then y + (-8) = -2 → y = +6.

Thus vanadium goes from +5 to +3, and sulfur goes from +4 to +6. This matches option D.

Key Takeaways

  • Always assign oxidation numbers using the known rules: oxygen is -2 (except in peroxides), hydrogen is +1, and the sum equals the charge on the species.
  • In redox reactions, the oxidation number of one species increases (oxidation) and another decreases (reduction). Here sulfur is oxidised and vanadium is reduced.

Common Mistakes

  • Forgetting that the sum of oxidation numbers in a polyatomic ion equals the ion's charge, not zero.
  • Assigning oxygen as -1 instead of -2.
  • Confusing the oxidation number of sulfur in SO2\text{SO}_2 (which is +4) with that in SO3\text{SO}_3 or sulfate.

Things to Be Careful About

  • Read the table carefully: the columns are vanadium (from), vanadium (to), sulfur (from), sulfur (to).
  • Double-check the charge on each ion before assigning oxidation numbers.
  • The reaction is not balanced, but that does not affect oxidation number assignment.
Techniques used
assign oxidation numbers to elements in polyatomic ionsdetermine oxidation state changes in redox reaction

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