Question

As carried out in Section A, a 20 cm³ aliquot of a 0.0575 mol dm^-3 vanadium(V) solution wasreduced by excess zinc metal to an unknown oxidation state (x). The resulting lavender (lilac) coloured solution was then fully reoxidised by 32.5 cm³ of a 0.0199 mol dm^-3 KMnO₄solution.

Calculate the experimental value for the unknown oxidation state x (to two decimal places)

Answer 1

Concentration of V⁵⁺ = 0.0575 mol dm^-3

Volume of aliquot = 20 cm³ = 0.02 dm³

Moles of V⁵⁺ = 0.0575 * 0.0   = 0.00115

Concentration of KMnO₄ solution = 0.0199 mol dm^-3

Volume of solution = 32.5 cm³ = 0.0325 dm³  

Moles of KMnO₄ = 0.0199 * 0.0325

= 0.00064675

When V⁵⁺ is reduced to +x odidation state using Zn:

2V⁵⁺ + 2(5 - x)e- 2V^x+  

(5 - x)Zn     (5 - x)Zn²⁺   + 2(5 - x)e-
---------------------------------------------------------------
2V⁵⁺ +   (5 - x)Zn   2V^x+ + (5 - x)Zn²⁺

Now,

When V^x+ is odidized by KMnO₄ solution.

3V^x+      3V⁵⁺ + 3(5 - x)e-

(5 - x)MnO₄^-  +   3(5 - x)e-     (5 - x)MnO₂
---------------------------------------------------------------------------------------
3V^x+   +   (5 - x)MnO₄^-       3V⁵⁺ +   (5 - x)MnO₂

As 3 moles V^x+ are oxidised by (5 - x) moles of KMnO₄  

So, 0.00115 moles V^x+ will be oxidised by = ((5 - x) / 3) * 0.00115 moles of KMnO₄  

But we have 0.00064675 moles of KMnO₄

Therefore,

((5 - x) / 3) * 0.00115 = 0.00064675

x = 3.31