This discussion is now closed.

Check out other Related discussions

- A-Level chemistry question
- A-Level Group 7 Chemistry help
- Etha
- help with workings pls
- A LEVEL CHEMISTRY - Amount of substances question help
- alevel chemistry question
- Chem enthalpy change
- Redox question help
- ionic equations
- chemistry aqa aslevel inorganic question
- Nucleophilic substitution - Need help
- Help with ionic equation chemistry
- Chemistry balancing redox equations
- Is this answer from MathsWatch right?
- AQA GCSE Chemistry Paper 1 (Higher Tier) 8462/1H - 27 May 2022 [Exam Chat]
- A-level Chemistry
- WJEC Unit 1 Maths As Level Discussion 15 May 2024
- AQA A-Level Chemistry Paper 1 (7405/1) - 10th June 2024 [Exam Chat]
- A level chemistry help
- Need Help on an Electrochem Q

Right i know how to combine 2 half equations to an ionic equation how do i get from that to a full equation? Any help please?

Just pair up the ions to make salts?

Say for example they give you this:

2C02 + 2e -----> C2O4 2- (Equation 1)

MnO4- +8H+ +5e ---> Mn2+ +4H2O (Equation 2)

Reverse the first equation to get C2O4 2- ----> 2CO2 + 2e (Equation 3)

tHEN BALANCE THE e, this is done by multiplying equation 3 by 5 and equation 2 by 2

To get 5C2O4 2- ---->10CO2 + 10E

2MnO4- + 16H+ + 10e ----> 2Mn2+ + 8H20

Then add both together, to get the overall equation:

5c2042- + 2Mn04- +16 H+ -----> 10C02 + 2Mn2+ + 8H20

Please note when adding the H+ may cancel out, just like this equation

And do not add the electrons in the final equation

2C02 + 2e -----> C2O4 2- (Equation 1)

MnO4- +8H+ +5e ---> Mn2+ +4H2O (Equation 2)

Reverse the first equation to get C2O4 2- ----> 2CO2 + 2e (Equation 3)

tHEN BALANCE THE e, this is done by multiplying equation 3 by 5 and equation 2 by 2

To get 5C2O4 2- ---->10CO2 + 10E

2MnO4- + 16H+ + 10e ----> 2Mn2+ + 8H20

Then add both together, to get the overall equation:

5c2042- + 2Mn04- +16 H+ -----> 10C02 + 2Mn2+ + 8H20

Please note when adding the H+ may cancel out, just like this equation

And do not add the electrons in the final equation

Why is the first equation reversed can someone explain please i understand the rest.

you have to pair up a reduction with an oxidation to make an equal number of electrons on both sides when you add the equations together.

What you are effectively saying is that equation 1 must provide the electrons that are going to be used in equation 2 (or vice versa)

The way to find out which of the half equations is a reduction and which is an oxidation requires use ofelectrode potentials.

Put together a pair of half equations - one reduction (electrons on the left hand side)and one oxidation (electrons on the right hand side)

now check out E= E(red) - E(ox)

...where E(red) is the electrode potential of the half equation undergoing reduction and E(ox) is the electrode potential of the half equation undergoing oxidation.

If the value for Eº given is positive and greater than +0.3 (this value differs in different courses some say +0.4) then the reaction is spontanous when the two half equations are added together.

If you are happy with the equations this way round (ie Eº is > +0.3) then multiply each equation to make the number of electrons on the left in one half equation equal to the number of electrons on the RHS of the other equation and simply add the two equations together, cancelling out the electrons in the process.

To make a full equation WITH balancing ions you have to add in all of the original balancing ions from the species that you originally used - this process is fundamentally worthless as all the other ions are spectator ions and not involved in the reaction.

But, if you are desperate to do so for example:

after constructing the ionic equation you may get...

MnO4(-) + 5Fe2+ + 8H+ --> Mn2+ + 5Fe3+ + 4H2O

the MnO4(-) was added using KMnO4 so that makes one K+ ion

the Fe2+ was added using FeSO4 so that makes 5SO4(2-) ions

the H+ weas added using H2SO4 so that makes 4 more SO4(2-) ions

so now our equation looks like:

KMnO4 + 5FeSO4 + 4H2SO4 --> MnSO4 + 5/2Fe2(SO4)3 + 4H2O + 1/2K2SO4

multiply through by two to remove fractions

2KMnO4 +10FeSO4 +8H2SO4 --> 2MnSO4 + 5Fe2(SO4)3 + 8H2O + K2SO4

pretty useless but correct!

What you are effectively saying is that equation 1 must provide the electrons that are going to be used in equation 2 (or vice versa)

The way to find out which of the half equations is a reduction and which is an oxidation requires use ofelectrode potentials.

Put together a pair of half equations - one reduction (electrons on the left hand side)and one oxidation (electrons on the right hand side)

now check out E= E(red) - E(ox)

...where E(red) is the electrode potential of the half equation undergoing reduction and E(ox) is the electrode potential of the half equation undergoing oxidation.

If the value for Eº given is positive and greater than +0.3 (this value differs in different courses some say +0.4) then the reaction is spontanous when the two half equations are added together.

If you are happy with the equations this way round (ie Eº is > +0.3) then multiply each equation to make the number of electrons on the left in one half equation equal to the number of electrons on the RHS of the other equation and simply add the two equations together, cancelling out the electrons in the process.

To make a full equation WITH balancing ions you have to add in all of the original balancing ions from the species that you originally used - this process is fundamentally worthless as all the other ions are spectator ions and not involved in the reaction.

But, if you are desperate to do so for example:

after constructing the ionic equation you may get...

MnO4(-) + 5Fe2+ + 8H+ --> Mn2+ + 5Fe3+ + 4H2O

the MnO4(-) was added using KMnO4 so that makes one K+ ion

the Fe2+ was added using FeSO4 so that makes 5SO4(2-) ions

the H+ weas added using H2SO4 so that makes 4 more SO4(2-) ions

so now our equation looks like:

KMnO4 + 5FeSO4 + 4H2SO4 --> MnSO4 + 5/2Fe2(SO4)3 + 4H2O + 1/2K2SO4

multiply through by two to remove fractions

2KMnO4 +10FeSO4 +8H2SO4 --> 2MnSO4 + 5Fe2(SO4)3 + 8H2O + K2SO4

pretty useless but correct!

- A-Level chemistry question
- A-Level Group 7 Chemistry help
- Etha
- help with workings pls
- A LEVEL CHEMISTRY - Amount of substances question help
- alevel chemistry question
- Chem enthalpy change
- Redox question help
- ionic equations
- chemistry aqa aslevel inorganic question
- Nucleophilic substitution - Need help
- Help with ionic equation chemistry
- Chemistry balancing redox equations
- Is this answer from MathsWatch right?
- AQA GCSE Chemistry Paper 1 (Higher Tier) 8462/1H - 27 May 2022 [Exam Chat]
- A-level Chemistry
- WJEC Unit 1 Maths As Level Discussion 15 May 2024
- AQA A-Level Chemistry Paper 1 (7405/1) - 10th June 2024 [Exam Chat]
- A level chemistry help
- Need Help on an Electrochem Q

Latest

Trending

Last reply 2 weeks ago

Why does temperature affect Kc only and not concentration or pressure?Last reply 1 month ago

Hey!! Can anyone please tell me how we can make 0.01% solution from 5% stock solution