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Oxidation Number Method

H2SO4 + SO2 +...

Question

$H_{2} S O_{4} + S O_{2} + N a_{2} c r o_{4} \to C r_{2} (S O_{4})_{3} + N a_{2} S O_{4}$
Can u balance the give equation with oxidation method

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Solution

$^{1} H_{2}^{+ 6} S O_{4}^{+ 2} +^{+ 4} S O_{2}^{- 2} +^{+ 4} S O_{2}^{- 2} + N a_{2}^{1} C r_{2}^{+ 6} O_{7}^{+ 2} ⇌^{1} N a_{2}^{+ 6} S O_{4}^{- 2} +^{+ 3} C r_{2}^{6} (S O_{4})_{3}^{- 2} +^{+ 1} H_{2} O^{2}$

Oxidation: $^{+ 4} S O_{2}^{- 2} ⟶^{1} {N a_{2}}^{+ 6} S O_{4}^{- 2} + 2 e^{-}$
Reduction: $6 e^{-} +^{1} {N a_{2}}^{+ 6} {C r_{2}}^{- 2} O_{7} ⟶^{+ 3} {C r_{2}}^{6} (S O_{4}^{- 2})_{3}$
$O : 3 S O_{2} ⟶ 3 N a_{2} S O_{4} + 6 e^{-}$
$R : N a_{2} C r_{2} O_{7} + 6 e^{-} ⟶ C r_{2} (S O_{4})_{3}$
$3 S O_{2} + N a_{2} C r_{2} O_{7} ⇌ 3 N a_{2} S O_{4} + C r_{2} (S O_{4})_{3}$

$\Rightarrow 3 S O_{2} + N a_{2} C r_{2} O_{7} + H_{2} S O_{4} ⇌ N a_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$

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