Write balanced net ionic equation for the following reaction in acidic solution.
$S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)$

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Solution

The given unbalanced equation on solving with ion electron method, we get the balanced equation.
The balanced equation is
${6 S}_{2} O_{3}^{- 2} + {C r}_{2} O_{7}^{- 2} + 14 H^{+} ⇌ {3 S}_{4} O_{6}^{- 2} + 2 {C r}^{+ 3} + {7 H}_{2} O$

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Similar questions

H^{+}

S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)

S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)

I O_{3}^{-} (a q) + R e (s) \to R e O_{4}^{-} (a q) + I^{-} (a q)

S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)

H^{+}

H^{+}

S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)

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