HelpContactFor teachers and institutes
>_StemPy Lab
Mock TestsClassesGroups

>_StemPy Lab

Learn. Solve. Code the solutions. Test yourself.

Live examsFree mock exams, timed and rankedEveryone starts together, and your score and rank come the moment it ends.See the exams →

Learn

  • Physics
  • Chemistry
  • Mathematics
  • Biology
  • Python Programming
  • Statistics & Data Science
  • Classes
  • Groups

Practice

  • Problems
  • Quizzes
  • Mock Tests
  • Prepare for an Exam
  • Question Sets
  • Exams
  • Certificates
  • Python Labs
  • AI Practice Generator
  • Student-Created Tests

Community

  • Discussions
  • Resources
  • Notes
  • Blog

Account

  • Dashboard
  • My Content
  • Profile
  • Log in
© 2026 senLearn. All rights reserved.HelpAboutContactPrivacy PolicyTerms of UseRefund Policy
  1. Problems
  2. /Copper reacts with concentrated nitric acid and gives copper(II) nitrate, brown nitrogen dioxide…
EasyChemistryHigh School (11–12)Redox Reactions › Balancing Redox Equations and Redox TitrationsNumericalA-LevelAPIB

Copper reacts with concentrated nitric acid and gives copper(II) nitrate, brown nitrogen dioxide…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Copper reacts with concentrated nitric acid and gives copper(II) nitrate, brown nitrogen dioxide gas and water. The skeleton equation is Cu+HNO3→Cu(NO3)2+NO2+H2O\text{Cu} + \text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + \text{NO}_2 + \text{H}_2\text{O}Cu+HNO3​→Cu(NO3​)2​+NO2​+H2​O. Balance it by the oxidation number method.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

Need a reminder? The key ideaShowHide

The total rise in oxidation number equals the total fall: electrons lost equal electrons gained. Match the changes first. Then balance the atoms that did not change, charge, hydrogen and oxygen.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Oxidation numbers. Copper goes from 000 to +2+2+2, a rise of 2. Nitrogen goes from +5+5+5 in HNO3\text{HNO}_3HNO3​ to +4+4+4 in NO2\text{NO}_2NO2​, a fall of 1.

Step 2. Match. One copper atom loses 2 electrons. Each NO2\text{NO}_2NO2​ gains 1. So 2 nitrogen atoms are reduced for each copper atom: Cu+HNO3→Cu(NO3)2+2NO2+H2O\text{Cu} + \text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + \text{H}_2\text{O}Cu+HNO3​→Cu(NO3​)2​+2NO2​+H2​O.

Step 3. Nitrogen. Two nitrate ions stay in the salt, and two nitrogen atoms go into NO2\text{NO}_2NO2​. So 4 molecules of HNO3\text{HNO}_3HNO3​ are needed.

Step 4. Hydrogen and oxygen. Four hydrogen atoms on the left give 2 water molecules. Oxygen: 12 on the left, and 6+4+2=126 + 4 + 2 = 126+4+2=12 on the right.

Answer: Cu+4HNO3→Cu(NO3)2+2NO2+2H2O\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}Cu+4HNO3​→Cu(NO3​)2​+2NO2​+2H2​O. The coefficients asked for are 4, 2 and 2.

Log in to save this for later
← All Problems
← Name the kind of reaction, and say whether it is a redox reaction: (a) 2Mg + O₂ → 2MgO; (b) 2HgO →…Zinc reacts with very dilute nitric acid… →
More onRedox Reactions › Balancing Redox Equations and Redox Titrations

Discussion (0)

Log in to join the discussion.

Loading…

Learn it in

  1. The oxidation number methodBalancing Redox Equations and Titrations

More on this topic

  1. A 10.0 mL sample of bleach solution is added to excess potassium iodide in acid…Medium
  2. A 25.0 mL sample of diluted hydrogen peroxide is acidified and titrated with 0.0180 mol/L potassium…Medium
  3. A 20.0 mL sample of iron(II) sulfate solution needs 16.0 mL of 0.0150 mol/L acidified potassium…Easy
Hot nitric acid oxidises arsenic(III) sulfide to arsenic acid and sulfate ions, and is itself…Hard
  • In basic solution, hydrogen peroxide oxidises chromium(III) hydroxide to chromate ions: Cr(OH)₃ +…Medium
  • All problems on this topic →