Write balanced net ionic equation the following reaction in acidic solution-S-2O-3-2-aq-Cr-2O-7-2-aq- rightarrow S-4O-6-2-aq-Cr-3-aq-

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Toppr Admin · Accepted Answer

The given unbalanced equation on solving with ion electron method- we get the balanced equation-The balanced equation is6S2O-x2212-23-Cr2O-x2212-27-14H-x21CC-3S4O-x2212-26-2Cr-3-7H2O

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

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

Write balanced net ionic equation for the following reaction in acidic solution.S2O2−3(aq)+Cr2O2−7(aq)→S4O2−6(aq)+Cr3+(aq)

The given unbalanced equation on solving with ion electron method, we get the balanced equation.The balanced equation is6S2O−23+Cr2O−27+14H+⇌3S4O−26+2Cr+3+7H2O

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

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