Chemistry 9701/13 — May/June 2013
Cambridge AS Level · Multiple Choice (AS Level) · answer key with instant marking and worked solutions
Topics Atoms, Molecules and Stoichiometry · States of Matter · Hydrocarbons · Introduction to Organic Chemistry · Electrochemistry · Atomic Structure · +12 more
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In the redox reaction shown, how do the oxidation states of vanadium and sulfur change?
Options
| vanadium (from) | vanadium (to) | sulfur (from) | sulfur (to) | |
|---|---|---|---|---|
| A | +1 | +3 | 0 | -2 |
| B | +1 | +3 | +4 | +6 |
| C | +5 | +3 | 0 | -2 |
| D | +5 | +3 | +4 | +6 |
Working
Assign oxidation numbers:
- In : each O is -2, total -4, charge +1, so V = +5.
- In : V = +3.
- In : each O is -2, total -4, molecule neutral, so S = +4.
- In : 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
D
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:
We need to find the oxidation numbers of vanadium and sulfur on both sides and see how they change.
Approach
- Determine the oxidation number of vanadium in and in .
- Determine the oxidation number of sulfur in and in .
- Compare the changes and select the correct option.
Step-by-Step Reasoning
Vanadium in :
- 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 :
- The ion has a charge of +3, so V = +3.
Sulfur in :
- Each oxygen is -2, so two oxygens contribute -4.
- The molecule is neutral, so S = +4.
Sulfur in :
- 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 (which is +4) with that in 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.
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