Assume the following reaction is carried out in acidic medium.
$6 e^{-}$ + $C r_{2} {O_{7}}^{2 -}$ $⟶$ 2 ${C r}^{3 +}$ + $7 H_{2} O$

How many $H^{+}$ do we need to add to the LHS to balance the H atoms?

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Solution

The correct option is D
14

Count number of Hydrogen on right hand side 7 $\times$ 2 = 14
Hence, add 14 $H^{+}$ ions to the side which is deficient.
$C r_{2} {O_{7}}^{2 -} + 6 e^{-} + 14 H^{+} ⟶$ 2{Cr}^{3+} + 7H_2O\)

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Similar questions

Assume the following reaction is carried out in acidic medium.
$6 e^{-} + C r_{2} {O_{7}}^{2 -} ⟶ 2 {C r}^{3 +} + 7 H_{2} O$

How many $H^{+}$ do we need to add to the LHS to balance the H atoms?

Q. Consider the reaction :

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

What is the quantity of electricity in coulombs needed to reduced

1 m o l

C r_{2} O_{7}^{2 -}

To balance H in Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) + 7H₂O (in acidic medium), we will

Q. Balance the following equation by ion electron method:

C r_{2} O_{7}^{2 -} + C_{2} H_{4} O + H^{+} \to 2 C r^{3 +} + C_{2} H_{4} O_{2} + H_{2} O

Q. For the equation in acidic medium:

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 coefficient of

H^{+}

in the balanced reaction by oxidation number method will be :

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Assume the following reaction is carried out in acidic medium. 6e− + Cr2O72− ⟶2 Cr3+ + 7H2O How many H+ do we need to add to the LHS to balance the H atoms?

The correct option is D 14 Count number of Hydrogen on right hand side 7 ×2 = 14 Hence, add 14 H+ ions to the side which is deficient. Cr2O72−+6e−+14H+⟶ 2{Cr}^{3+} + 7H_2O\)

Assume the following reaction is carried out in acidic medium.
$6 e^{-}$ + $C r_{2} {O_{7}}^{2 -}$ $⟶$ 2 ${C r}^{3 +}$ + $7 H_{2} O$

How many $H^{+}$ do we need to add to the LHS to balance the H atoms?

The correct option is D
14

Count number of Hydrogen on right hand side 7 $\times$ 2 = 14
Hence, add 14 $H^{+}$ ions to the side which is deficient.
$C r_{2} {O_{7}}^{2 -} + 6 e^{-} + 14 H^{+} ⟶$ 2{Cr}^{3+} + 7H_2O\)