Redox EquationsWatch

#1
The question is as follows: (I will just include the redox equations needed)

-> Look at the following half equations. Use them to create the redox equations stated below ;
- Ce(3+) -> Ce(4+) + e(-)
- MnO4(-) + 8H(+) + 5e(-) -> Mn(2+) + 4H2O

a) Oxidation of Ce(3+) by H(+)/MnO4(-)

Would appreciate a step-by-step guide on this or an answer which I can work with for the rest of the questions!

Cheers,
Andrew
0
6 years ago
#2
So, basically it is asking you to combine two half equations. So, you just add everything on reactants side with that on products side but doing that means you have more electrons on reactants than on products but you can't have independent electrons. So, you need to balance electrons in both half equations before adding them together.

Spoiler:
Show
1. First balance electrons in half equations.
- [Ce3+ -> Ce4+ + e-]*5 => 5Ce3+ -> 5Ce4+ + 5e-
- MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

2. Now the electrons are balanced, just add the two half equations.
- 5Ce3+ + MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O + 5Ce4+ + 5e-

3. You have equal electrons on both sides so you can cancel them out (think of it as algebra).
- 5Ce3+ + MnO4- + 8H+ -> Mn2+ + 4H2O + 5Ce4+
1
#3
(Original post by swanderfeild)
So, basically it is asking you to combine two half equations. So, you just add everything on reactants side with that on products side but doing that means you have more electrons on reactants than on products but you can't have independent electrons. So, you need to balance electrons in both half equations before adding them together.

Spoiler:
Show
1. First balance electrons in half equations.
- [Ce3+ -> Ce4+ + e-]*5 => 5Ce3+ -> 5Ce4+ + 5e-
- MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

2. Now the electrons are balanced, just add the two half equations.
- 5Ce3+ + MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O + 5Ce4+ + 5e-

3. You have equal electrons on both sides so you can cancel them out (think of it as algebra).
- 5Ce3+ + MnO4- + 8H+ -> Mn2+ + 4H2O + 5Ce4+
Thank you very much for the quick response! Turns out I got the right answer; ideally, I should post my answers too!

Thanks!
0
#4
(Original post by swanderfeild)
So, basically it is asking you to combine two half equations. So, you just add everything on reactants side with that on products side but doing that means you have more electrons on reactants than on products but you can't have independent electrons. So, you need to balance electrons in both half equations before adding them together.

Spoiler:
Show
1. First balance electrons in half equations.
- [Ce3+ -> Ce4+ + e-]*5 => 5Ce3+ -> 5Ce4+ + 5e-
- MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

2. Now the electrons are balanced, just add the two half equations.
- 5Ce3+ + MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O + 5Ce4+ + 5e-

3. You have equal electrons on both sides so you can cancel them out (think of it as algebra).
- 5Ce3+ + MnO4- + 8H+ -> Mn2+ + 4H2O + 5Ce4+
So in the case of "Oxidation of H2O2 by H+/MnO4(-)
-Half equations of : 1)MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O
2) H2O2 --> 2H(+) + O2 + 2e(-)

Would the answer come out to be;
- 5H2O2 + 2MnO4(-) + 16H(+) --> 10H(+) + 5O2 + 2Mn(2+) + 8H2O
0
6 years ago
#5
(Original post by andrewruss)
Thank you very much for the quick response! Turns out I got the right answer; ideally, I should post my answers too!

Thanks!
You are welcome.

(Original post by andrewruss)
So in the case of "Oxidation of H2O2 by H+/MnO4(-)
-Half equations of : 1)MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O
2) H2O2 --> 2H(+) + O2 + 2e(-)

Would the answer come out to be;
- 5H2O2 + 2MnO4(-) + 16H(+) --> 10H(+) + 5O2 + 2Mn(2+) + 8H2O
Yes but you should simplify it further by subtracting 10H+ on both sides so same species isn't on both sides of the equation.
0
#6
(Original post by swanderfeild)
You are welcome.

Yes but you should simplify it further by subtracting 10H+ on both sides so same species isn't on both sides of the equation.
Would that leave us with
5H2O2 + 2MnO4(-) + 6H(+) --> 5O2 + 2Mn(2+) + 8H2O

Isn't the previous version of the answer (
5H2O2 + 2MnO4(-) + 16H(+) --> 10H(+) + 5O2 + 2Mn(2+) + 8H2O) correct? It has equal charges on either side.
0
2 years ago
#7
this is true but you can simplify further by subtracting elements that appear on both sides for example : 5H202 2Mn04 --> 8H20 ....Because water is on both sides you can subtract by the smaller number so subtract by 5 on both sides to get 2Mn04 ---> 3H20.
0
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