Balance Cr2O7-2 + S2O3-2= Cr+3 + SO4-2

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Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

Q. Balance the following chemical equation:

C r_{2} O_{7}^{- 2} (a q) + S O_{3}^{- 2} (a q) \to C r^{+ 3} (a q) + S O_{4}^{- 2} (a q)

(Acidic medium)

When $N a N O_{2}$ is oxidised by $K_{2} C r_{2} O_{7}$ , the important products formed are

Q. Arrange the following in increasing flocculating power for coagulation of positive sol.

[F e (C N)_{6}]^{- 4}, P O_{4}^{- 3}, S O_{4}^{- 2}, C l^{-}

Q. Iodine oxidises

S_{2} O_{3}^{- 2}

ion into:

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