0.108 g of finely divided copper was treated with an excess of ferric sulphate solution untill Cu was completely dissolved. The solution after the addition of excess dil `H_2SO_4`, required `33.7mL` of 0.1 N `KMnO_4` for complete oxidation. Find the equation which represents the reaction between metallic copper and ferric sulphate solution. `(Cu=63.7,Fe=56)`

Explanation

33.7 " mL of " `0.1 N KMnO_4=3.37 m" Eq of "KMnO_4`
`=0.00337 " Eq of "KMnO_4`
`[Fe^(2+)` ions are formed on reduction of `Fe_2(SO_4)_3` by Cu]
" Eq of "`Cu(0.108)/(63.5)=0.0017`
0.0017 " Eq of "Cu produces 0.00337 " Eq of "`Fe^(2+)`
1 Eg of Cu produces `=(0.00337)/(0.0017)approx2" Eq of "Fe^(2+)`
`approx2 " mol of "Fe^(2+)`
`approx2 " mol of "FeSO_4`
Hence, the equation is
`Cu+Fe_2(SO_4)_3toCuSO_4+FeSO_4`