Chapter 23 Redox And Electrode Potentials Flashcards
General recap about redox
Redox reactions is when BOTH oxidation and reduction take place
The thing oxidised increases redox number, loses electrons and is the REDUCING AGENT
Thing reduced is the OXIDISING AGENT
How to balance and MAKE A HALF EQUATION if you know what species before becomes after
For example
MNO4- —> MN2+
1) balance atoms
2) CHECK OXYGENS, oxygens become h2o on the OTHER SIDE
now it’s MNO4- —> Mn2 + + 4H2O
3) now balance the HYDROGENS, these becomes H+ on left
MNO4 - + 8H+ —> Mn2+ + 4H2O
4) finally balance the CHARGE, and make adjustments by adding electrons
So here charge on right is 2 and on left is 7, so add 5 electrons on left will balance equation
Common half equations we should know (what do they become because you can work the rest out:
MNO4-
Cr2o7 2-
Becomes Mn2+
Becomes Cr 3 +
Once have two half equations how to balance?
Multiply electrons, cancel and cancel species on both sides, combine to make overall equation
How to BALANCE AN EQUATION USING REDOX NUMBERS
+ when to?
FINAL CHECK
If balancing looks long + there is hydorgen on left that is split up into different things on the right, then might need to use this method
1) write the equation
2) IDENTIFY WHATS BEING OXIDISED AND WHATS BEING REDUCED . Write down the OVERAL amount it’s being oxiside and reduces by
3) we know that the total increase in oxidation number MUST equal the total decrease .
So if s increases by 6 and the other decreases by 1, we must put a 6 around the thing that decreases to balance oxidation numbers.
4) check if oxidation numbers balanced, now see if anything else needs balancing and finish
5) AS A FINAL CHECK MAKE SURE THE CHARGES ARE RHE SAME ON BOTH SIDES
Key point about equations, what must be balanced with oxidation number and reduction?
(The rule that increase in oxidation = ?)
The increase of oxidation number = the decrease
So if oxidation increased by 6, then it must also decrease by 6 overall elsewhere
Total should be 0!
Sometimes you need to predict species to complete half equations
What are common species to be predicted?
H+ , OH- , H2O
When balancing equations do the charges have to be 0 on both sides or what?
Just the SAME charge whatever it is, make sure to check
When balancing, and there’s one species that becomes two, how can you manipulate the net redox gain
So like 4HCl becomes Pbcl2 and cl2, what’s the redox change?
Here -4 becomes -2 and 0
So overall, add 2 both sides, the change is -2 to 0 so just an INCREASE OF 2
Use this technique to help balance !
How to do the MNO4 2- to Mn2+ redox titration?
1) what are the colour changes?
2) what else MUST WE ADD
As MNo4 is being REDUCED (+7 to +2), need something that will be OXIDISED
= common substance is Fe 2+ being oxidised to Fe 3+
1) put the manganite in the burette And the iron 2 + ion solution in beaker .
2) ADD DILUTE H2SO4, to provide the H+ ions needed for the redox reactions
3) this reaction is SELF INDICATING, as the Mn goes form purple to colourless . Therefore the first drop you see that is permanent pale pink, that means you hit the endpoint as all the Fe2+ has been reacted
Again what must you add to the MNO4 2- redox titration in order to make it work
An EXCESS of H2so4 to provide the H+ ions needed
If it asks for percentage of whole molecule or just iron what do you do?
If it’s whole molecule purity use fm of it all
Of it’s just IRON, then only use Fm of IRON!
What happens in the thisulfste iodine redox (colour wise)
What to do for the end point
Here the thisulfste is being reduced , whereas Solid iodine is being oxidised
As a result, the iodine is becoming orange ti colourless, and the titration stops once the solution goes COLOURLESS
However the colour gets fainter and fainter and can be some discrepancy to when it becomes colourless.
2) === NEAR THE END POINT, when it vehicles a pale STRAW colour, add some starch.
- starch will make solution go black in presence of iodine, and then when iodine leaves it becomes colourless again
- this helps you see the end point
Why must we add the starch near the END OF THE TITRARIOM when a pale straw colour seen only?
This is because the starch iodine complex is very strong, and if we add it at the start there is a chance some of the iodine just sticks the starch FOREVER, and is missed out in the titration
As a result we add it near the end when the concentration of iodine is LOW , to avoid this irreversible stick
So what’s the idea behind using thisulfste and iodine redox. How does it let us find the concentration of something else?
YOU START your reaction by reacting the thing you want to identify with POTASSIUM IODIDE (a source of iodine IONS)
These will react and form a certain quantity of IODINE SOLID
You then react this iodine solid with thisulfste in the reaction
- and work backwards to find the concentration= it’s a TWO STEP REDOX REACTION
