Redox reactions Flashcards
The price of copper is increasing as supplies of high-grade ores start to run out. The mineral covellite (CuS), found in low-grade ores, is a possible future source of copper.When copper is extracted from covellite, a reaction occurs between copper(II) sulfide and nitric acid to form a dilute solution of copper(II) sulfate.Balance the equation for this reaction.3CuS(s) + ……HNO3(aq)——-> ……CuSO4(aq) + ……NO(g) + ……H2O(I)Give the oxidation state of nitrogen in each of the following. HNO3 ……………………………………….NO ………………………………………….Deduce the redox half-equation for the reduction of the nitrate ion in acidified solution to form nitrogen monoxide and water.Deduce the redox half-equation for the oxidation of the sulfide ion in aqueous solution to form the sulfate ion and H+(aq) ions.
3CuS(s) + 8HNO3(aq) ——–>3CuSO4(aq) + 8NO(g) + 4H2O(l)(+) 5 (+) 24H+ + NO3– + 3e– ———>2H2O + NOS2– + 4H2O ————>SO42– + 8e– + 8H+
Use your knowledge of metal reactivity to state and explain a low-cost method for the extraction of copper from a dilute aqueous solution of copper(II) sulfate.Write the simplest ionic equation for the reaction that occurs during this extraction process.
add scrap / recycled / waste iron (or steel) to the aqueous solutionthe iron is a more reactive metal OR Fe is a better reducing agentCu2+ /copper ions are reduced / gain electrons OR Cu2+ +2e–> CuOR copper / Cu is displaced by FeFe + Cu2+ ——–>Fe2+ + Cu
Barium metal reacts very quickly with dilute hydrochloric acid, but it reacts more slowly with water.Write an equation for the reaction of barium with water.
Ba + 2H2O —->Ba(OH)2 + H2
A solution containing barium ions can be used to show the presence of sulfate ions in an aqueous solution of sodium sulfate.Write the simplest ionic equation for the reaction that occurs and state what is observed.Simplest ionic equation Observation
Ba2+ + SO4(2-) ——->BaSO4White precipitate / solid
State one use of barium sulfate in medicine.Explain why this use is possible, given that solutions containing barium ions are poisonous.
Barium meal / barium swallow / barium enemaOR used in X-rays OR to block X-rays OR X-ray contrastmedium OR CT scansBaSO4 / barium sulfate is insoluble (and therefore not toxic)
Give the formula of a Group 2 metal hydroxide used in agriculture.
Ca(OH)2 OR Mg(OH)2
Identify a sodium halide that does not undergo a redox reaction when added as a solid to concentrated sulfuric acid.
NaF or sodium fluoride ORNaCl or sodium chloride
Chlorine gas reacts with cold dilute sodium hydroxide solution to form sodium chloride and another chlorine-containing compound, X.Give the formula of X.
NaClO OR NaOCl
Give the formula of the substance responsible for the orange colour when chlorine gas is bubbled through an aqueous solution of sodium bromide.
Br2
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid. Give the formula for each of the following in this reaction.Formula of the solid reduction product ……………………………………………………………………Formula of the oxidation product ……………………………………………………………………………
S OR S8 OR S2I2
Reactions that involve oxidation and reduction are used in a number of important industrial processes.Iodine can be extracted from seaweed by the oxidation of iodide ions.In this extraction, seaweed is heated with MnO2 and concentrated sulfuric acid.Give the oxidation state of manganese in MnO2
MnO2 (+) 4
Write a half-equation for the reaction of MnO2 in acid to form Mn2+ ions and water as the only products.
MnO2 +4H+ +2e– ——–> Mn2+ +2H2O
In terms of electrons, state what happens to the iodide ions when they are oxidised.
Iodide ion(s) is/are oxidised because they have lost electron(s)
Chlorine is used in water treatment. When chlorine is added to cold water it reacts to form the acids HCl and HClOThe following equilibrium is established.Cl2(aq) + H2O(I)  H+(aq) + Cl–(aq) + HClO(aq) Give the oxidation state of chlorine in Cl2 and in HClO
Cl2 0 HClO (+) 1
Deduce what happens to this equilibrium as the HClO reacts with bacteria in the water supply. Explain your answer (2)
Equilibrium will shift / move to the rightto oppose the loss of HClO (or replace the HClO that has reacted)
Concentrated sulfuric acid is reduced when it reacts with solid potassium bromide.Concentrated sulfuric acid is not reduced when it reacts with solid potassium chloride. Write the two half-equations for the following redox reaction.
2Br― ——–> Br2 + 2e―and2H+ + SO42- + 2e― ——-> SO2 +2H2O
Write an equation for the reaction of solid potassium chloride with concentrated sulfuric acid.
KCl + H2SO4 ——-> KHSO4 + HCl
Explain why chloride ions are weaker reducing agents than bromide ions (2)
Relative size of ions:Chloride ions are smaller than bromide ions OR chloride ion electron(s) are closer to the nucleus OR chloride ion has fewer (electron) shells / levels OR chloride ion has less shielding (or converse for bromide ion)Strength of attraction for electron being lost:Outer shell / level electron(s) OR electron(s) lost from a chloride ion is more strongly held by the nucleus compared with that lost from a bromide ion (or converse for bromide ion)
Nitric acid is manufactured from ammonia in a process that involves several stages.In the first stage, ammonia is converted into nitrogen monoxide and the following equilibrium is established.4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ∆H=–905kJmol–1The catalyst for this equilibrium reaction is a platinum–rhodium alloy in the form of a gauze. This catalyst gauze is heated initially but then remains hot during the reaction.In terms of redox, state what happens to the ammonia in the forward reaction.
Oxidation
Suggest a reason why the catalyst must be hot.
Any one from• to provide / overcome activation energy • to provide the minimum energy to make the reaction go / start
Suggest a reason why the catalyst remains hot during the reaction.
The reaction is exothermic OR releases heat (energy)
State how a catalyst increases the rate of a reaction(2)
Catalysts provide an alternative route / pathway of lower activation energy
Nitrogen monoxide, NO, is formed when silver metal reduces nitrate ions, NO3-, in acid solution. Deduce the oxidation state of nitrogen in NO and in NO3-
.NO3-We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states equal to the charge of the ion, which is equal to -1. The oxidation state of Nitrogen is x.x + 3(-2) = -1x - 6 = -1x = -1 +6 = +5.NOWe know that the Oxygen has an oxidation state of -2. The sum of the oxidation states equal to 0. The oxidation state of Nitrogen is x.x -2 = 0x = +2
Write a half equation for the reduction of NO3- ions in acid solution to form nitrogen monoxide and water.
NO3- + 4H+ +3e- —–> NO + 2H2O
Write a half-equation for the oxidation of silver metal to Ag+(aq)ions
Ag —–> Ag+ + e-
Hence, deduce an overall equation for the reaction between silver metal and nitrate ions in acid solution
NO3- + 4H+ +3Ag —–> NO + 2H2O + 3Ag+
Explain, with reference to electron transfer, what is meant by the term oxidising agent.
The oxidising agent accept electrons
In the presence of a strong acid, the IO3- ion is a powerful oxidising agent. The half-equation (ion-electron equation) for this process in shown below:IO3-(aq)+6H+(aq)+5e- —–> ½ I2(aq) + 3H2O(l)Under acidic conditions, IO3- will oxidise iodide ions to iodine.i- Deduce the oxidation numbers of iodine in, IO3-, I- and I2ii- Write an ionic equation to show the reaction between aqueous solution of KIO3 and KI under acidic conditions
iIO3-(aq)+6H+(aq)+5e- —–> ½ I2(aq) + 3H2O(l).IO3- We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to the charge of the ion which is here -1. The oxidation state of I is x.x + 3(-2) = -1x - 6 = -1x = -1 + 6x = +5.I- Oxidation state is equal to the charge of the ion. Therefore the oxidation state of I is -1.I2 Uncombined elements have oxidation state equal to 0.iiIO3- + 6H++ 5I- —–> 3H2O + 3I2
Define reduction in terms of electrons.
Reduction is gain of electrons.
Define an oxidising agent in terms of electrons.
Oxidising agents accept electrons.
In aqueous solution, bromine oxidises sulphur dioxide, SO2 , to sulphate ions,SO42-Deduce that oxidation state of sulfur in SO2 and in SO42-
.SO2We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to 0. The oxidation state of Sulfur is x.2(-2)+ x = 0-4 + x = 0x = +4 .SO42- We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to the charge of the ion which is -2. The oxidation state of Sulfur is x.x + 4 (-2) = -2x - 8 = -2x = -2 + 8x = +6
In terms of electron transfer, what does the reducing agent do in a redox reaction?
a reducing agents give away electrons - they are electron donors.
Uncombined elements have oxidation state 0.
Uncombined elements have oxidation state 0.
educe the oxidation state of nitrogen in each of the following compounds.(i)NCl3(ii)Mg3N2(iii)NH2OH
.NCl3We know that the Chlorine has an oxidation state of -1. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.x +3(-1) = 0x -3 = 0x = +3.Mg3N2We know that the Magnesium has an oxidation state of +2. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.3(+2)+ 2(x) = 0+6 + 2x = 02x = -6x = -6 2x = -3.NH2OHWe know that the Hydrogen has an oxidation state of +1 and Oxygen of -2. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.x + 2(+1)- 2 + 1 = 0x + 2 - 2 + 1 = 0x +1 = 0x = -1
Lead(IV) oxide,PbO2, reacts with concentrated hydrochloric acid to produce chlorine, lead(II) ions,Pb2+, and water(i) Write a half equation for the formation of Pb2+ and water from PbO2 in the presence of H+ ions
i+4 +2PbO2 + H+ ——> Pb2+ + H2O.PbO2We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to 0. The oxidation state of Pb is x.x + 2(-2)= 0x - 4 = 0x = +4PbO2 + 4H+ + 2e- ——> Pb2+ + 2H2O
Lead(IV) oxide,PbO2, reacts with concentrated hydrochloric acid to produce chlorine, lead(II) ions,Pb2+, and water(i) Write a half equation for the formation of Pb2+ and water from PbO2 in the presence of H+ ions
i+4 +2PbO2 + H+ ——> Pb2+ + H2O.PbO2We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to 0. The oxidation state of Pb is x.x + 2(-2)= 0x - 4 = 0x = +4PbO2 + 4H+ + 2e- ——> Pb2+ + 2H2O2Cl- ——> Cl2 +2e-
Lead(IV) oxide,PbO2, reacts with concentrated hydrochloric acid to produce chlorine, lead(II) ions,Pb2+, and water(i) Write a half equation for the formation of Pb2+ and water from PbO2 in the presence of H+ ions
i+4 +2PbO2 + H+ ——> Pb2+ + H2O.PbO2We know that the Oxygen has an oxidation state of -2. The sum of the oxidation states is equal to 0. The oxidation state of Pb is x.x + 2(-2)= 0x - 4 = 0x = +4PbO2 + 4H+ + 2e- ——> Pb2+ + 2H2O
Write a half-equation for the formation of chlorine from chloride ions
2Cl- ——> Cl2 +2e-
Deduce the oxidation state of nitrogen in each of the following compounds.(i)NCl3(ii)Mg3N2(iii)NH2OH
.NCl3We know that the Chlorine has an oxidation state of -1. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.x +3(-1) = 0x -3 = 0x = +3.Mg3N2We know that the Magnesium has an oxidation state of +2. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.3(+2)+ 2(x) = 0+6 + 2x = 02x = -6x = -6 2x = -3.NH2OHWe know that the Hydrogen has an oxidation state of +1 and Oxygen of -2. The sum of the oxidation states is equal to 0. The oxidation state of Nitrogen is x.x + 2(+1)- 2 + 1 = 0x + 2 - 2 + 1 = 0x +1 = 0x = -1
