OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 3
1 mark · Medium difficulty · Multiple Choice
Deduce the correct stoichiometric balancing numbers for silver and water in the redox reaction between silver and concentrated nitric acid.
Practise this questionQuestion
Question text
3 The unbalanced equation for the reaction of silver with concentrated nitric acid is shown below.
….. Ag(s) + ….. NO –(aq) + ….. H+(aq) → ….. Ag+(aq) + .…. NO(g) + ..… H O(l)
Which numbers for Ag and H2O will balance the equation?
Ag(s) H2O(l)
A 1 2
B 2 3
C 3 1
D 3 2
Your answer [1]
Mark scheme
Show the mark scheme
3 D 1 AO2.6
How to answer it
Balancing Ionic Redox Equations
What this question tests
This question assesses your ability to determine oxidation numbers, construct half-equations, and use electron-transfer logic to balance a complex ionic redox equation in acidic conditions.
Balancing Silver with Concentrated Nitric Acid
The unbalanced equation is given as:
..... Ag(s) + ..... NO₃⁻(aq) + ..... H⁺(aq) → ..... Ag⁺(aq) + ..... NO(g) + ..... H₂O(l)
Which numbers for Ag(s) and H₂O(l) will balance the equation?
✅ Correct Answer: D (Ag = 3, H₂O = 2)
Option D correctly balances both the electron transfer (oxidation states) and the atomic/charge conservation, yielding 3 moles of silver and 2 moles of water.
💡 Key Knowledge
- Oxidation state changes: Ag changes from 0 to +1 (loses 1 electron). Nitrogen in NO₃⁻ changes from +5 to +2 in NO (gains 3 electrons).
- Electron balancing: To balance electrons, you need 3 Ag atoms for every 1 NO₃⁻ reduced to NO.
- Acidic conditions: H⁺ ions and H₂O molecules are used to balance oxygen and hydrogen atoms after balancing the main redox elements.
🧠 Exam Technique
Instead of blindly guessing coefficients from the options, construct the balanced equation systematically using oxidation numbers or half-equations. Once fully balanced, read off the required coefficients for Ag and H₂O .
❌ Common Errors
- Balancing only atoms by inspection without tracking oxidation number changes.
- Confusing the oxidation state of nitrogen in nitrate ( +5 ) with nitrogen dioxide or other nitrogen species.
- Failing to account for the stoichiometric ratio required for electron exchange (the 3:1 ratio between Ag and NO).
📐 Step-by-Step Derivation
- Identify oxidation numbers: Ag goes from 0 to +1. N in NO₃⁻ goes from +5 to +2 in NO .
- Set up electron transfer: Each Ag loses 1 e⁻. Each N gains 3 e⁻. Therefore, you need 3 Ag for every 1 NO .
- Place preliminary coefficients: Put a 3 in front of Ag , Ag⁺ , NO₃⁻ , and NO :
3 Ag + 1 NO₃⁻ + ..... H⁺ → 3 Ag⁺ + 1 NO + ..... H₂O
- Balance Oxygen and Hydrogen:
- Left side has 3 oxygens (from NO₃⁻). Right side has 1 oxygen (in NO). We need 2 more oxygens on the right, so add a coefficient of 2 in front of H₂O : 2 H₂O .
- This introduces 4 hydrogens on the right ( 2 × H₂ ). Balance this by putting a 4 in front of H⁺ on the left.
- Final Balanced Equation:
3 Ag(s) + 1 NO₃⁻(aq) + 4 H⁺(aq) → 3 Ag⁺(aq) + 1 NO(g) + 2 H₂O(l)
- Match with options: Coefficient for Ag = 3, coefficient for H₂O = 2. This matches option D.
Topics
Module 2: Foundations in chemistry · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.