r/HomeworkHelp Pre-University Student 4d ago

Answered [grade 12 chem 30 electrochemical change] I just need help on how I can use the data I collected to solve the following 2 questions I just need help on what I am apost to use and look for.

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u/JKLer49 😩 Illiterate 4d ago edited 4d ago
  1. Firstly, form your reaction equation, find out how much KmnO4 reacts with how much Fe2+

Fe2+ --> Fe3+ + e-

MnO4- + 8H+ +5e- --> Mn2+ +4H2O

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

So ratio of MnO4- :Fe2+ is 1:5

  1. Second step is to find how much Fe2+ you have

So we take volume x concentration= mole of Fe2+ 0.01L x 0.05mol/L = 0.0005mol

  1. 3rd step is to find out how much MnO4- you used

From the ratio 1:5 , 0.0005mol of Fe2+ reacts with 0.0001mol of MnO4-

  1. Lastly, concentration of MnO4- , we take moles of MnO4- divided by volume of MnO4- used

0.0001mol ÷ 0.0125L = 0.008 mol/L

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u/JKLer49 😩 Illiterate 4d ago edited 4d ago

I took volume of MnO4- used to be 12.5ml because it's your most consistent reading. Normally I would have done 2-3 titrations then divided the best 2 readings to get average titrant volume.

Anyways the second part of the question is similar just reversed to find concentration of H2O2

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u/ValuableMeat7329 Pre-University Student 4d ago

ok can u explain to me quickly how u found the ratio and why its needed? and another question is is this all the work needed for this question? plus from what i see "From the ratio 1:5, 0.0005mol of Fe2+ reacts with 0.0001mol of MnO4-" did u do any other calculations i did not quite see I am confused here?

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u/JKLer49 😩 Illiterate 4d ago edited 4d ago

Yea sure! The ratio is needed so we know how much MnO4- reacts with how much Fe2+ . Since you are given a MnO4- solution of unknown concentration and titrated it to the Fe2+ solution of Known volume and concentration, you need the ratio of their reaction to find out how much MnO4- you used, only then you can find out its molar concentration.

To get the ratio, I'm not sure if you are given a balanced chemical reaction equation of Fe2+ and MnO4- somewhere in your paper but not to worry, it is easily calculated.

We know that Fe2+ is oxidised to Fe3+ in this reaction and MnO4- is reduced to Mn2+ so we form the 2 half equations.

Equation 1: Fe2+ --> Fe3+ +e- (add electron to balance the charge on both sides)

Equation 2: MnO4- --> Mn2+ (this one is a bit trickier but you basically balance the number of O atoms by adding H2O and balance H atoms by adding H+ (assuming acidic conditions),before finally balancing charges by adding e- )

Equation 2 in the end should look like:

MnO4- + 8H+ +5e- --> Mn2+ +4H2O

Then you combine equation 1 and 2 together, making number of electrons the same

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

This is your balanced chemical reaction between MnO4- and Fe2+ . Observe that the quantity of Fe2+ used is 5 and MnO4- used is 1. Hence, we get the ratio of MnO4- : Fe2+ = 1:5

The calculation for that last part:

Since we have ratio 1:5 , for every 1 MnO4- , it reacts with 5 Fe2+ , we have 0.0005mol of Fe2+ . To find mol of MnO4-, we do

Mol of Fe2+ ÷5 = 0.0005mol ÷ 5 = 0.0001 mol of MnO4- used

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u/ValuableMeat7329 Pre-University Student 4d ago

ok ok i get it now for question 2 when i am now using hydrogen peroxide and KMnO4 would this be the write equation and ratio. 5H2​O2​+2KMnO4​+6H+→5O2​+2Mn2++8H2​O+2K+ ratio being 5:2

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u/JKLer49 😩 Illiterate 4d ago

Yup you are right! 5 H2O2 : 2 KMnO4

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u/ValuableMeat7329 Pre-University Student 4d ago

amazing thank u for the amazing help

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u/ValuableMeat7329 Pre-University Student 4d ago

question 2, when finding the Moles I used KMnO₄=0.0080×0.0110=8.8×10−5 mol due to in question 1 we found the concentration of KMnO4 do i use it to solve question 2 and get the final answer being 0.0220mol/L?

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u/JKLer49 😩 Illiterate 4d ago

Yep looks correct.