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### Course: Chemistry library > Unit 14

Lesson 1: Oxidation-reduction reactions- Oxidation and reduction
- Oxidation state trends in periodic table
- Practice determining oxidation states
- Unusual oxygen oxidation states
- Balancing redox equations
- Oxidizing and reducing agents
- Disproportionation
- Worked example: Balancing a simple redox equation
- Worked example: Balancing a redox equation in acidic solution
- Worked example: Balancing a redox equation in basic solution
- Redox titrations
- Oxidation–reduction (redox) reactions

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# Redox titrations

A redox titration is a titration in which the analyte and titrant react through an oxidation–reduction reaction. As in acid–base titrations, the endpoint of a redox titration is often detected using an indicator. Potassium permanganate (KMnO₄) is a popular titrant because it serves as its own indicator in acidic solution. Created by Jay.

## Want to join the conversation?

- I understand that the Manganese ion has been reduced by gaining electrons that has been lost by the iron ion. But I don't quite get the role of the sulfuric acid here. Is it that it's providing free hydrogen ions that transfer the electrons between the Manganese and iron ions??(29 votes)
- The sulfuric acid provides the H+ for the reaction. MnO4- only reacts in an acidic solution because the oxygen in the MnO4- react with the H+ to form water. Without the acidified solution, the oxygen would remain with the Mn and no redox reaction would occur.(56 votes)

- Why do you use only 10 mL instead of the total 30 mL to calculate the molarity?(16 votes)
- Because we're interested in the initial concentration of the Fe2+, and we only had 10 mL of that.(32 votes)

- Why does the Fe^2+ turn into Fe^3+ when reacted with MnO4^-?(4 votes)
- This may be a simplistic argument, but the Mn in MnO₄⁻ is in the +7 oxidation state.

No atom "wants" to give up its electrons; after all, it takes almost 3000 kJ/mol to remove the electron from Fe²⁺ and convert it to Fe³⁺.

But Mn " is desperate" to get electrons to reach a lower oxidation state.

Like the bully it is, it grabs electrons from as many Fe²⁺ ions as it can find to get to a +2 state.

It takes 10Fe²⁺ ions to produce enough electrons to reduce 2 Mn(VII) to 2Mn²⁺ ions.

The Mn might take even more electrons from Fe, but it takes a whopping 5300 kJ/mol to convert Fe³⁺ to Fe⁴⁺, and few things are able to supply that much energy.(18 votes)

- When you were finding the mol for MnO4 why did you use 20ml instead 10ml? I am a bit confuse at this point. Cheers(4 votes)
- He used 20 mL of 0.02 M KMnO₄ from the buret for the titration, so he wanted to calculate the number of moles (0.0004 mol).

Then he knew that he had 5 × 0.0004 mol of Fe²⁺ in the original sample.

But the original sample had a volume of only 10 mL, so he used 10 mL to calculate its molarity.(12 votes)

- Why Potassium Permanganate? where did they get that?(4 votes)
- KMnO₄ is just a standard oxidizing agent that they used to oxidize Fe²⁺ to Fe³⁺.

KMnO₄ is a great oxidizing agent for redox titrations because, once all the Fe²⁺ has reacted, one more drop of KMnO₄ produces a permanent purple colour.(11 votes)

- the potassium (K+) in the potassium permangenate isnt going to mess up the titration outcome with its positive charge?(6 votes)
- Nope, it just acts as a spectator ion and does not react with anything because potassium (K), an alkali metal, is soooo reactive that it remains in solution the way it is.(3 votes)

- This might sound really dumb but where did the equation come from? Did you need a redox table?(4 votes)
- Not at all a stupid question :P It is simply a balanced equation (earlier lessons) knowing what happens with the ions based on experimental observation:

(MnO4)- + (Fe)2+ --> (Mn)2+ + (Fe)3+

See for yourself by balancing the equation I stated and you should get the same as in the video.(3 votes)

- if i use the mv shortcut which side does the 5 come and why?(3 votes)
- If you use the MV method, then you divide each side by the coefficient for each side.

(M(MnO4-) x V(MnO4-) )/1 = (M(Fe2+) x V(Fe2+))/5

As to why, it's simpler for you to just try it with the numbers in the video. You will find it works.(4 votes)

- How do I predict these reactions when products are not given??(3 votes)
- some reactions have predictable outcomes. for example, when we combust a hydrocarbon (in O2), the products are typically water and carbon dioxide.(1 vote)

- What if there are no oxygens in the reaction? What does the hydrogen bond to? Also, how did they find that it was the iron reacting to form the color?(2 votes)
- The titration is done in water. There are always oxygen atoms present.

They already know the colours of the ions.

Fe²⁺ is a very pale green.

Fe³⁺ is a very pale brown.

MnO4⁻ is a deep purple.

Mn²⁺ is a very pale pink, almost colourless.(3 votes)

## Video transcript

- [Voiceover] We've already seen how to do an acid-base titration. Now let's look at a redox titration. Let's say we have a solution containing iron two plus cations. We don't know the concentration of the iron two plus cations, but we can figure out the concentration by doing a redox titration. Let's say we have 10
milliliters of our solution, and let's say it's an acidic solution. You could have some
sulfuric acid in there. In solution, we have iron two plus cations and a source of protons from our acid. To our iron two plus solution, we're going to add some
potassium permanganate. In here, we're going to have some potassium permanganate, KMnO4. Let's say the concentration of our potassium permanganate is .02 molar. That's the concentration
that we're starting with. Potassium permanganate is, of course, the source of permanganate anions, because this would be K plus and MnO4 minus. Down here, we have a source
of permanganate anions. We're going to drip in the potassium permanganate solution. When we do that, we're going to get a redox reaction. Here is the balanced redox reaction. If you're unsure about how to balance a redox reaction, make sure to watch the video on balancing redox reactions in acid. Let's look at some oxidation
states really quickly so we can see that this
is a redox reaction. For oxygen, it would be negative two. We have four oxygens, so negative two times four is negative eight. Our total has to add up
to equal negative one. For manganese, we must have a plus seven, because plus seven and negative eight give us negative one. Manganese has an oxidation
state of plus seven. Over here, for our products, we're going to make Mn two plus. Manganese two plus cation in solution, so the oxidation state is plus two. Manganese is going from an oxidation state of plus seven to plus two. That's a decrease or a reduction in the oxidation state. Therefore, manganese is being reduced in our redox reaction. Let's look at iron two plus. We have iron two plus as one of our reactants here. That means the oxidation
state is plus two. For our products, we're making iron three plus, so an oxidation state of plus three. Iron is going from plus two to plus three. That's an increase in the oxidation state. Iron two plus is being oxidized in our redox reaction. As we drip in our potassium permanganate, we're forming our products over here. These ions are colorless in solution. As the permanganate reacts, this purple color disappears and we should have colorless, we should have a colorless solution. Let's say we've added a
lot of our permanganate. Everything is colorless. But then we add one more drop, and a light purple color persists. Everything was clear, but then we add one drop of permanganate and then we get this light purple color. This indicates the
endpoint of the titration. The reason why this is the endpoint is because our products are colorless. So if we get some purple color, that must mean we have some unreacted, a tiny excess of unreacted
permanganate ions in our solution. That means we've completely reacted all the iron two plus that we originally had present. So we stop our titration at this point. We've reached the endpoint. We've used a certain volume of our potassium permanganate solution. Let's say we finished down here. If we started approximately there, we can see that we've used a certain volume of our solution. Let's say it took 20 milliliters. We used up 20 milliliters of our potassium permanganate solution to completely titrate our iron two plus. Our goal was to find the
concentration of iron two plus. If we're going to find the concentration of iron two plus, we could figure out how many moles of permanganate were necessary to completely react
with our iron two plus. We could figure out moles from molarity and volume. Let's get some more room down here. We know that molarity is equal to moles over liters. The molarity of permanganate is .02. We have .02 for the concentration
of permanganate ions. Moles is what we're solving for. It took us 20 milliliters
for our titration, which we move our decimal
place one, two, three, so we get .02 liters. So solve for moles. .02 times .02 is equal to .0004. So we have .0004. This is how many moles of permanganate were needed to completely react with all of the iron two plus that we originally had in our solution. It took .0004 moles of permanganate to completely react with our iron. All right. Next, we need to figure out how many moles of iron two plus that we originally started with. To do that, we need to use our balance redox reaction. We're going to look at the coefficients, because the coefficients
tell us mole ratios. The coefficient in front
of permanganate is a one. The coefficient in front of iron two plus is a five. If we're doing a mole
ratio of permanganate to iron two plus, permanganate would be a one and iron would be a five. So we set up a proportion here. One over five is equal to ... Well, we need to keep permanganate in the numerator here. How many moles of
permanganate were necessary to react with the iron two plus? That was .0004. So we have .0004 moles of permanganate. X would represent how many
moles of iron two plus we originally started with. We could cross-multiply
here to solve for x. Five times .0004 is equal to .002. X is equal to .002. X represents the moles of iron two plus that we originally had present. We're almost done, because our goal was to
find the concentration of iron two plus cations. Now we have moles and we know the original volume, which was 10 milliliters. To solve for the concentration of iron two plus, we just take how many moles
of iron two plus we have, which is .002, so we have .002 moles of iron two plus. We started with a total volume of 10 milliliters, which is equal to .01 liters. So we have .01 liters here. And .002 divided by .01 is equal to .2. So this is equal to .2 molar. That was the original concentration of iron two plus ions in solution. You could have used the MV is equal to MV
equation and modified it, because our ratio isn't one to one here. That's another way to do it. But I prefer to actually sit down and do these calculations and think about exactly what's happening.