A-Level chemistry question

A student finds an old bottle of vanadium chloride which has lost part of its label and the student is unsure of the oxidation state and hence the formula of the substance.The label states that the concentration of the solution is 0.0764 moldm-3.A student wants to establish the charge of the vanadium ions (V* ) so they can deduce the vanadium chlorides formula.They know that the solution can be titrated with MnOr, in the presence of Ht to form vanadate (V) ions, VOs, in a redox titration as the solutions colour changes on addition of Mn04. The manganate (VII) ions (MnO4) convert into manganese (1) ions in the process.The student titrates 25cm3 of the 0.0764 moldm 3 Vanadium solution with 38.10cms of 0.0200moldm-3 KMn04.a) Write out the half equation for the reduction of MnO,.b) Calculate the moles of electrons gained by manganese during the titration.c) From part b) and the moles of Vx , deduce the value for X and the formula of the vanadium chloride.d) Write the half equation for the reaction of VX to VOs.e) Using your answers to part (a) and (d) construct the overall equation for the reaction.If you have not been able to correctly work out the value of x create an overall equation using the half equation below. This is not the correct answer to part (d)V2 3H20 → VO3^ 6H 3е-Please help me. I am stuck on what to do

A student finds an old bottle of vanadium chloride which has lost part of its label and the student is unsure of the oxidation state and hence the formula of the substance.The label states that the concentration of the solution is 0.0764 moldm-3.A student wants to establish the charge of the vanadium ions (V* ) so they can deduce the vanadium chlorides formula.They know that the solution can be titrated with MnOr, in the presence of Ht to form vanadate (V) ions, VOs, in a redox titration as the solutions colour changes on addition of Mn04. The manganate (VII) ions (MnO4) convert into manganese (1) ions in the process.The student titrates 25cm3 of the 0.0764 moldm 3 Vanadium solution with 38.10cms of 0.0200moldm-3 KMn04.a) Write out the half equation for the reduction of MnO,.b) Calculate the moles of electrons gained by manganese during the titration.c) From part b) and the moles of Vx , deduce the value for X and the formula of the vanadium chloride.d) Write the half equation for the reaction of VX to VOs.e) Using your answers to part (a) and (d) construct the overall equation for the reaction.If you have not been able to correctly work out the value of x create an overall equation using the half equation below. This is not the correct answer to part (d)V2 3H20 → VO3^ 6H 3е-Please help me. I am stuck on what to do

Thanks for the help. Based on what you said, I think the equation for reduction of MnO4- is MNO4- + 5e- + 8H+ —-> Mn2+ + 4H2O

That is correct.
How are you finding the rest of the question?

For part b) my answer for the moles of electrons gained was 0.00381 moles. Part c) onwards is where I’m a bit stuck

Alright. My answer for that was 0.00191 moles

Okay, so for that I divided the moles of electrons transferred by the moles of V^+ ions and got 1.99, which I rounded to 2.

Oxidation state of V in VO3^- is +5. So oxidation state must have been +3 if it had to lose two electrons. So does that mean X is 3?

I agree with your answer.

So can you now write an ionic equation (again, using H2O and H^+ to balance it) for the oxidation of V^3+ to VO3^-?

Okay so for this I got 5V^3+ + 2MnO4- + 16H+ ——-> 5VO3^- + 2Mn^2+ + 8H2O

It might help to first derive an ionic half-equation showing just the oxidation of V^3+ to VO3^- and then try combining half-equations afterwards.

My ionic equation for V^3+ was: V^3+ ——> VO3^- + 2e- Maybe this was the equation I got incorrect? I understand the reduction ionic equation for MnO4^-. But I think the V^3+ oxidation ionic equation was the one that messed up my combined overall redox equation

Have you tried balancing that half-equation by putting water on the left hand side and H^+ on the right hand side?

I think it is V^3+ + 3H2O ——> VO3^- + 2e- + 6H+

I agree. How might you use that and MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O to write an overall equation with no electrons on either side?

The overall equation I got was 5V^3+ + 15H2O. + 2MnO4^- + 16H+ ——-> 5VO3^- + 3OH^+ + 2Mn^2+ + 8H2O