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Chemistry - Chemical Reactions - Displacement reactions
Terms in this set (11)
A piece of copper wire placed in a solution of silver nitrate undergoes a displacement reaction.
- Describe any observations that would be made
- Link these observations to the chemical species involved
- Explain why the displacement reaction occurs
- Write a balanced ionic equation for this reaction
Cu(s) + 2Ag+(aq) ---> Cu2+(aq) + 2Ag(s)
- The copper wire is a shiny brownish-red and the silver nitrate is a colourless solution. When the copper wire is added to the solution of silver nitrate the copper wire begins to disappear and overtime white solid is formed on the copper wire. Overtime the colourless solution of silver nitrate changes to a blue coloured solution.
- The white solid formed is silver metal and the change in colour of the solution from colourless to blue refers to the removal of the silver ions in the solution and the adding of copper ions into the solution. The disappearance of the copper wire is also referring to the adding of copper ions into the solution.
- Displacement occurs because copper metal is more reactive than silver metal. So the copper metal is able to lose two electrons and displaces the less reactive silver out of the solution. While the silver ions are removed from the solution and it gains the lost electrons from the copper atoms to become atoms in the solid that we see forming.
An experiment in the laboratory involves placing an iron nail in copper sulfate solution and a copper nail in iron sulfate solution. Both nails are cleaned with sandpaper, placed in each solution, and the test tubes are stoppered and left overnight.
- Describe any observations that would be made the next day
The iron nail in the copper sulfate will have a brown deposit on it and will slowly dissolve and the blue copper sulfate solution will turn to pale green. The test tube with with the copper nail in the iron sulfate will remain unchanged and the nail will still be a brownish-red colour and the solution of iron sulfate will be pale green.
Identify the type of reaction that occurs and give a reason for your choice.
- Balanced ionic equation for ONE reaction that occurs
This is a displacement reaction as iron is able to displace the copper ions from the solution because iron metal is more reactive than copper metal, as iron is more reactive than copper.
Balanced symbol equation
Fe(s) + Cu2+(aq) ---> Fe2+(aq) + Cu(s)
Cleaned pieces of metals iron (Fe), lead (Pb) and tin (Sn) were placed solutions of iron(II) sulfate (FeSO4), lead nitrate (Pb(NO3)2) and tin sulfate (SnSO4).
Metal - FeSO4 Sol - Pb(NO3)2 Sol - SnSO4 Sol
Iron - not tested - reaction - reaction
Lead - no reaction - not tested - no reaction
Tin - no reaction - reaction - not tested
Use the information provided in the table to place tin (Sn) in the activity series
When the iron metal is placed in the solution of lead nitrate and tin sulfate a reaction occurred. This means that iron metal is more reactive than lead metal and tin metal so is able to lose two electrons and force the lead ion/tin ion to accept these two electrons and revert to its element form (neutral). (iron displaces both tine ions and lead ions from the solution) Therefore iron is above tin and lead in the activity series. Lead is less reactive than iron because when placed in the solution of iron sulfate no reaction occurred. It is also less reactive than tin as no reaction occurred when lead was placed in the solution of tin sulfate, so it will not displace Fe2+ or Sn2+. Therefore lead will be below tin and iron in the activity series. When tin is placed in the solution of iron sulfate no reaction occurred because tin metal is not as reactive than iron metal and therefore unable to displace Fe2+ from the solution. However a reaction occurred when tin was places in lead nitrate, therefore tin is more reactive than lead metal as it was able to displace Pb2+ from the solution. Therefore tin is less reactive than iron but more reactive than lead so it will be between iron and lead on the activity series.
identify the type of reaction that is occurring (displacement) and justify your choice by referring to ONE specific example from the reactions.
(Magnesium and copper nitrate)
- Identify the reaction you have selected
- Describe any observations that would be made in the selected reaction, and link your observations to the reactants and products involved in the reaction
- Identify the type of reaction occurring, and justify your choice
- Write a balanced symbol equation for your reaction
- I have selected the reaction where magnesium is added to the solution of copper nitrate.
- Magnesium is a shiny grey metal and copper nitrate is a blue solution, overtime the magnesium metal will slowly disappear and a brown deposit of copper will from on the magnesium. The blue copper nitrate solution will fade into a colourless solution as the copper ions are displaced from the solution resulting in a solution of magnesium nitrate.
- This is a displacement reaction since the magnesium metal has displaced the copper ions from the solution. This was able to occur because magnesium metal is more reactive than copper metal and is therefore able to lose two electrons and force the copper ions to gain these two electron and revert to a copper element (neutral). This is why the magnesium metal will start to disappear and the atoms making up the metal will become Mg2+ ions and go into the solution. The copper nitrate solution will go from a blue solution to a colourless solution due to the Cu2+ becoming Cu atoms and being deposited on the magnesium metal
Mg(s) + Cu2+(aq) ---> Mg2+(aq) + Cu(s)
Explain why magnesium nitrate solution does not react with any of the three metal zinc, copper and lead.
The colourless solution of magnesium nitrate doesn't react with zinc, copper and lead because magnesium metal is higher in the activity series and is therefore a more reactive metal than zinc, copper and lead. So zinc, copper and lead aren't able to lose electrons and force the magnesium ion to gain them and displace the magnesium ions from the solution, so no reaction would occur
A zinc rod was placed in a solution of copper sulfate and another zinc road was placed in a solution of magnesium sulfate and both were left for one week
- Describe any observations that would be made for each beaker
- Identify the type of reaction occurring
- Explain any difference in the observations made in beaker 1 and beaker 2 by linking your observations with the type of reaction occurring and the reactants and the products involved
- Write a balanced equation
The zinc rod will slowly disappear (decrease in mass) and a brown deposit of copper will form on the surface of the zinc rod. Overtime the blue copper sulfate solution will become colourless.
No reaction will occur so there will be no change and the grey zinc rod will stay the same in the colourless solution magnesium sulfate.
- Displacement reaction
- In beaker 1, zinc metal is more reactive than copper metal according to the activity series and therefore is able to displace the copper ions from the solution, so overtime the solution will go from blue to colourless (zinc ions being displaced in the solution). A brown deposit of copper will form on the zinc rod due to the copper being displaced from the solution. In beaker 2, there is no reaction because zinc metal is less reactive than magnesium metal and is therefore unable to displace the magnesium ions from the solution. So there will be no change to the zinc rod and the magnesium sulfate solution
Balanced symbol equation
Zn(s) + Cu2+(aq) ---> Zn2+(aq) + Cu(s)
In the laboratory the teacher made impure lead crystals by placing a metal in lead nitrate solution
- Choose a suitable metal for this reaction and justify your choice
- Balanced symbol equation
- Magnesium would be suitable for this experiment as it is higher than lead metal on the activity series so is therefore a more reactive metal. Because magnesium metal is more reactive than lead metal it is able to lose two electrons and force the lead ion to gain these two electrons and revert to a lead element, and displace the lead ions from the solution. A black deposit of lead will then form on the surface of the magnesium and this is the impure lead crystals
Balanced symbol equation
Mg(s) + Pb2+(aq) ---> Mg2+(aq) + Pb(s)
A piece of copper was added to a solution of silver nitrate in a beaker, and left for one day
- Identify the type of reaction occurring in the beaker
- Describe the observations occurring, and link them to the reactants and products involved
- Balanced ionic equation for the reaction occurring
- Copper is a shiny brownish-red metal and silver nitrate is a colourless solution. Overtime the copper metal will disappear and a grey solid deposit of silver will form on the surface of the copper. The colourless solution of silver nitrate turns blue as copper nitrate is formed
Balanced symbol equation
Cu(s) + 2Ag+(aq) ---> Cu2+(aq) + 2Ag(s)
A strip of silver-grey metal is know to be either silver or magnesium. Explain how the identity of the metal could be determined by adding copper sulfate solution to the strip of metal, and leaving it for one day
- Give any observations you would expect to see if the metal is silver and magnesium and link them to the relevant species present
- Explain why a chemical reaction may or may not occur, depending on whether the metal is silver or magnesium
If the unknown strip of silver-grey metal is added to the solution of copper sulfate and overtime the metal starts to disappear and the blue copper sulfate solution goes from blue to colourless and a brown deposit if formed on the metal then the silver-grey metal is magnesium. A reaction will occur between magnesium and copper sulfate solution because magnesium metal is more reactive than copper metal as it is higher in the activity series so the magnesium is able to lose two electron and force the copper ion to gain these two electrons and revert to the copper element (neutral). It is also able to displace the copper ions from the solution so it goes from blue to colourless due to magnesium ions being displaced in the solution. If the unknown strip of silver-grey metal is added to the copper sulfate solution and no reaction occurs then the metal is silver, as copper metal is more reactive than silver metal as it higher in the activity series so copper ions are unable to be displaced from the solution.
(Combined question with combination reaction, find the answer with the study set for combination reactions)
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