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# A level redox equation help

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1. This was actually a question from an old thread- just reposting it as I wanted some help with it.

A solution is made from, 50cm^3 ,iron (ii) chloride, diluted to 500cm^3. the solution was then titrated against both 0.02M potassium manganate, and 0.01677M potassium dichromate (separately). (after adding an excess of dilute sulpuric acid). 25cm^3 solution was used each time, 25.50cm^3 more manganate was needed to be added each time than dichromate. calculate the concentration of iron (ii) chloride in the original solution. before dilution)

Basically, this is my working so far:
1) (Cr2O7)2- + 6Fe2+ + 14H+---->6Fe3+ + 2Cr3+ +7H2O
2) MnO4- +5Fe2+ + 8H+----> 5Fe3+ + Mn2+ +4H2O
Then, I set: Volume of (Cr2O72-)= x, and the Volume of (MnO4-)= x+25.5
for reaction with Cr2O72-
moles of Cr2O72-= 0.01677 X (x/1000)= (0.01677x)/1000
moles of Fe2+ reacted in 25cm^3= 6(0.01677x)/1000= (0.10062x)/1000
moles of Fe2+ in 500cm^3= 20(0.10062x)/1000= (2.0124x)/1000
^^^^This is then the number of moles in 50cm^3 also (before dilution)

Then for the reaction with MnO4-:
moles of MnO4-= 0.02 X (x+25.5)/1000
moles of Fe2+ reacted in 25cm^3= moles above X5= 0.1(x+25.5)/1000
moles of Fe2+ in 500cm^3= moles above X 20= 2(x+25.5)/1000
^^^^This is also the number of moles in 50cm^3 (before dilution)

I then equated the two bolded values and came out with like x=4112.9....
Where did I go wrong???
2. charco could you possibly help out?
3. (Original post by thesmallman)
This was actually a question from an old thread- just reposting it as I wanted some help with it.

A solution is made from, 50cm^3 ,iron (ii) chloride, diluted to 500cm^3. the solution was then titrated against both 0.02M potassium manganate, and 0.01677M potassium dichromate (separately). (after adding an excess of dilute sulpuric acid). 25cm^3 solution was used each time, 25.50cm^3 more manganate was needed to be added each time than dichromate. calculate the concentration of iron (ii) chloride in the original solution. before dilution)

Basically, this is my working so far:
1) (Cr2O7)2- + 6Fe2+ + 14H+---->6Fe3+ + 2Cr3+ +7H2O
2) MnO4- +5Fe2+ + 8H+----> 5Fe3+ + Mn2+ +4H2O
Then, I set: Volume of (Cr2O72-)= x, and the Volume of (MnO4-)= x+25.5
for reaction with Cr2O72-
moles of Cr2O72-= 0.01677 X (x/1000)= (0.01677x)/1000
moles of Fe2+ reacted in 25cm^3= 6(0.01677x)/1000= (0.10062x)/1000
moles of Fe2+ in 500cm^3= 20(0.10062x)/1000= (2.0124x)/1000
^^^^This is then the number of moles in 50cm^3 also (before dilution)

Then for the reaction with MnO4-:
moles of MnO4-= 0.02 X (x+25.5)/1000
moles of Fe2+ reacted in 25cm^3= moles above X5= 0.1(x+25.5)/1000
moles of Fe2+ in 500cm^3= moles above X 20= 2(x+25.5)/1000
^^^^This is also the number of moles in 50cm^3 (before dilution)

I then equated the two bolded values and came out with like x=4112.9....
Where did I go wrong???
Your chemical equations are correct. I think there is something wrong with the values. If you work forward from a supposed answer of 0.51 M, the titres for permanganate and chromate would be nearly the same.
4. (Original post by TeachChemistry)
Your chemical equations are correct. I think there is something wrong with the values. If you work forward from a supposed answer of 0.51 M, the titres for permanganate and chromate would be nearly the same.
ohhhh okay cheers!

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