0.108 g of finely-divided copper was treated with an excess of ferric sulphate solution until copper was completely dissolved. The solution after the addition of excess dilute sulphuric acid required 33.7 mL of 0.1 N KMnO4 for complete oxidation. Find the equation which represents the reaction between metallic copper and ferric sulphate solution.
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Since, Cu will react with ferric sulphate to reduce Fe3+ to Fe2+, the reduced state of iron is further oxidized by KMnO4.
Thus, Meq. of KMnO4 used=Meq. of iron sulphate oxidized =Meq. of ferric sulphate used by Cu =Meq. of Cu
∴ Meq. of Cu=Meq. of KMnO4 used
(n63.6)0.108×=33.7×0.1
∴n=2
It is thus, clear that during reduction of Fe3+, Cu is oxidized to Cu2+.
Thus, reaction is as follows:
Cu+Fe2(SO4)3⟶CuSO4+FeSO4
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