balancing redox rxns....

[hoss19]

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has anyone seen DAT questions that ask you to fully balance a redox rxn? these problems take me at a minimum five minutes, i dont see how the test makers expect you to do these given the time sensitive nature of the test. any help, hints, feedback would be great!

 

[ELMerfud11]

I dont know about the actual DAT but i have definitely seen them in Practice tests ive been taking and in the destroyer. I heard though that they would jut ask for a half reaction they wouldnt ask you to balance out both sides. However i dont know if this is true.

 

[doc toothache]

has anyone seen DAT questions that ask you to fully balance a redox rxn? these problems take me at a minimum five minutes, i dont see how the test makers expect you to do these given the time sensitive nature of the test. any help, hints, feedback would be great!

Maybe you need to change your approach to balancing equations.

http://forums.studentdoctor.net/showpost.php?p=5831710&postcount=2

 

[tRNA]

Maybe you need to change your approach to balancing equations.

http://forums.studentdoctor.net/showpost.php?p=5831710&postcount=2

can you please elaborate a bit on this method?, i tried it with a different equation and it didn't work, so there must be something i am missing

 

[doc toothache]

can you please elaborate a bit on this method?, i tried it with a different equation and it didn't work, so there must be something i am missing

Can you post or pm the equation in question?

 

[DDSguyLA]

Hey,

They will NOT ask you to balance a full on redox on the actual test... The easiest way will take you a couple of minutes. They way I learned it takes me a good 4 minutes if I rush through it.

I HIGHLY DOUBT they will put it on a test. HOWEVER, they might put a redox and tell you that its in a basic of acidic solution and then three of the answers will have something that shouldn't like OH or maybe missing an OH, then you're stuck between only two, and you will see that, as soon as you start balancing, one other one will be eliminated.

This process takes me about 50 secs.


Good luck,

 

[tRNA]

this is the problem doc toothache

MnO4(-)+ Fe2+ --> Fe3+ + Mn2+

thanks

 

[Flipper405]

The only problem with redox rxns I can remember seeing is one where they ask you to identify the oxidizing and reducing agents... easy stuff

 

[klutzy1987]

this is the problem doc toothache

MnO4(-)+ Fe2+ --> Fe3+ + Mn2+

thanks

To do tis problem, first you break it into half reactions. Your two half reactions are
1.MnO4(-) -->Mn2+
then you balance his by first adding 4 oxygens to the left side to balance out oxygen. Oxygen is added via H2O. This results in an excess of 8H atoms on the left. To balance this we add 8H+ atoms on the right side. Now you have a balanced reaction besides for the charge. To balance the charge calculate the total charges on each side and add enough electrons to whichever side needs them in order to make the charges on both sides equal. In this case 5 electrons need to be added to the reactants in order to make the chrges equal. the final half reaction is 5e(-) + 8H(+) +MnO4(-) -->Mn(2+) + 4H2O
If you look you will see that both sides of the reaction are equal in charge and in molecules.

The other half reaction is
Fe(2+) --> Fe(3+) + 1e(-)

To cancel out the electrons we would multiply the second equation by 5 and then add up so that the 5 electrons one one side would cancel with the 5 electrons on the other side and there would be 0 electrons in the final equation.
After adding and cancelling out electrons and waters and H+ the final equation is

MnO4(-) + 5Fe(2+) +8H(+) --> Mn(2+) +4H2O + 5Fe(3+)

I hope this answers your question as it tooka while to type and i did it in my head so i cant be sure its coorrect but i think it is.

 

[doc toothache]

this is the problem doc toothache

MnO4(-)+ Fe2+ --> Fe3+ + Mn2+

thanks

A simple approach to balancing redox equation is as follows:

1. Write the oxidation state (valence of the elements involved) and determine the net gain and loss of electrons.


+7...............(+5e)................+2
MnO4- + Fe 2+----> Fe3+ + Mn2+
................ +2..(-1e)....+3

2. Since the gain of electrons must equal the loss of electrons, multiply the gain by the loss and the loss by the gain. In this case the gain is multiplied by 1 while the loss is multiplied by the gain of 5. These are the numbers you will be using for balancing the equation.

+7...............(+5e)................+2
MnO4- + Fe 2+----> Fe3+ + Mn2+
................ +2..(-1e)5....+3

and your equation (partial) will look as follows:

MnO4- + 5Fe 2+----> 5Fe3+ + Mn2+

(the spectator ions are balanced as needed)

 

[tRNA]

now i see it, doc toothache, so your way doesn't really balance the whole rxn, it just saves you the time of balancing the half rxns separately, then after your last step it becomes much fasster:
MnO4- + 5Fe 2+----> 5Fe3+ + Mn2+

you notice that there are 4oxygens on the left side so to balance that you add 4H2O to the right side, then you notice that there are 8Hs on the left side which are not on the left side so you add 8H+ to the left side to balance that out, charges and atoms are balanced.....and that's IT
yea much faster method....thanks

thanks Klutzy, i already knew how to do it the standard way just wanted the short cut

 

[doc toothache]

now i see it, doc toothache, so your way doesn't really balance the whole rxn, it just saves you the time of balancing the half rxns separately, then after your last step it becomes much fasster:
MnO4- + 5Fe 2+----> 5Fe3+ + Mn2+

you notice that there are 4oxygens on the left side so to balance that you add 4H2O to the right side, then you notice that there are 8Hs on the left side which are not on the left side so you add 8H+ to the left side to balance that out, charges and atoms are balanced.....and that's IT
yea much faster method....thanks

thanks Klutzy, i already knew how to do it the standard way just wanted the short cut

Not exactly a personal, but a way taught in your typical general chemistry textbook.

 

[hoss19]

i appreciate everyone's help with this

doc - this way is certainly MUCH quicker, thanks a ton. i notice that when the initial number of atoms are not the same on both sides of the given equation it is necessary to first balance those. one can then determine the coefficients necessary to multiply each of the oxidized / reduced species by followed by balancing w/ water, H+, and -OH (if needed). thanks again.