Chemical Changes (review)

1. Copper oxide can be formed by heating copper metal in air. The reaction is very slow compared with the same reaction using magnesium metal.

a) Explain why the reaction between copper and oxygen is an oxidation reaction.

Answer: Copper gains oxygen during the reaction.


b) Copper oxide contains Cu2+ and O2− ions. Give the formula of copper oxide.

Answer: CuO


c) Give the balanced symbol equation for the reaction between copper and oxygen to form copper oxide.

Answer: 2Cu + O2  →  2CuO


d) Describe the reaction that will take place when magnesium powder is heated with a sample of copper oxide. Include a word equation.

Answer: Magnesium is more reactive than copper. The reaction will be a displacement reaction. magnesium + copper oxide  →  magnesium oxide + copper


e) When solid copper oxide is warmed with dilute sulfuric acid (H2SO4 ), copper sulfate (CuSO4 ) and water are produced. This reaction can be used to produce pure, dry crystals of copper sulfate. Describe a method for producing pure, dry crystals of copper sulfate from solid copper oxide and dilute sulfuric acid.

Answer: Add dilute sulfuric acid to a beaker and add the copper oxide until it is in excess. Once the reaction has stopped filter out the excess copper oxide using filter paper and a funnel. Pour the solution of copper sulfate into an evaporating basin. Heat the solution until crystals start to form. Leave the solution to cool, filter out the crystals and allow them to dry.


2. The Group 1 metal sodium reacts quickly with cold water. The reaction produces a colourless gas and an alkaline solution.

a) Name the ion that causes the alkaline solution in the reaction described above.

Answer: hydroxide ion (OH)


b) Give the word equation for the reaction between sodium and water.

Answer: sodium + water  →  sodium hydroxide + hydrogen


c) During this reaction sodium atoms form sodium ions. Explain why this is an oxidation reaction.

Answer: Oxidation is the loss of electrons. Sodium atoms lose an electron to form the sodium ion.


d) A student reacted a small piece of lithium metal with water to produce a solution of lithium hydroxide (LiOH). They then added 10 cm3 of this solution to some dilute nitric acid (HNO3 ). Suggest the pH values for the lithium hydroxide solution and the nitric acid solution.

Answer: lithium hydroxide solution any value greater than 7 ; nitric acid any value between 1 and 3


e) Name the salt formed when lithium hydroxide is added to nitric acid.

Answer: Lithium nitrate


f) Give the ionic equation for the reaction between lithium hydroxide and nitric acid.

Answer: H+ + OH  →  H2O


3. Hydrochloric acid, HCl, is a strong acid. Dilute hydrochloric acid is commonly used in school laboratories to carry out simple chemical reactions. Concentrated hydrochloric acid is used in many industrial processes.

a) Explain the difference between the strength of an acid and how dilute or concentrated it is.

Answer: The strength of an acid describes how much the molecules ionise in water. How dilute or concentrated an acid is describes the number of acid molecules dissolved in a certain volume of water.


b) One sample of dilute hydrochloric acid has a pH of 2 while a sample of concentrated hydrochloric acid has a pH of –1. Determine how many times more concentrated the hydrogen ions are in the concentrated acid.

Answer: 1000 times more concentrated (for every 1 unit decrease in pH the concentration of H+ ions increases by a factor of 10)


c) A student was asked to carry out a titration experiment to determine the concentration of some dilute hydrochloric acid. They used 25.0 cm3 portions of sodium hydroxide solution with a concentration of 0.200 mol/dm3 and some phenolphthalein indicator. The equation for the reaction between sodium hydroxide and hydrochloric acid is KOH(aq) + HCl(aq) →  NaCl(aq) + H2O(l). The average volume of hydrochloric acid that was used in the titration experiment was 19.50 cm3. Calculate the concentration of the potassium hydroxide solution. Give your answer in mol/dm3 to three significant figures.

Answer: mol NaOH = concentration × volume = 0.200 mol/dm3 × (25.0 ÷ 1000) = 0.005 mol ;
mol HCl = mol NaOH (1:1 ratio in balanced equation) ;
concentration HCl = mol ÷ volume = 0.005 mol ÷ (19.5 ÷ 1000) = 0.25641 mol/dm3 = 0.256 mol/dm3


4. This is a question about magnesium and one of its compounds, magnesium nitrate. Magnesium reacts with acids to form a salt and hydrogen gas. Magnesium nitrate is a highly soluble salt.

a) Name the acid required to react with magnesium to form magnesium nitrate.

Answer: nitric acid


b) Magnesium nitrate contains the ions Mg2+ and NO3. Give the formula for magnesium nitrate.

Answer: Mg(NO3)2


c) During the reaction between magnesium and any acid both oxidation and reduction occur. The reaction is Mg + 2H+  →  Mg2+ + H2 . Explain why this reaction involves both oxidation and reduction reactions. Identify the substances being oxidised and reduced. Include half equations in your answer.

Answer: During the reaction Mg atoms lose electrons to form Mg2+ ions. Oxidation is the loss of electrons, so magnesium is oxidised: Mg → Mg2+ + 2e  or  Mg - 2e  →  Mg2+. Hydrogen ions (H+) gain electrons to form H2 molecules. Reduction is the gain of electrons, so hydrogen is reduced: 2H+ + 2e  →  H2


d) A solution of magnesium nitrate was electrolysed using inert graphite electrodes. Predict the products of the electrolysis reaction at the positive electrode (anode).

Answer: oxygen gas


e) Predict the products of the electrolysis reaction at the negative electrode (cathode).

Answer: hydrogen gas