To balance H in Cr 2 O 72- aq - Cr 3- aq -7 H 2 O in acidic medium- we willA- Add 14 H -on left sideB- Add 4 H -on left sideC- Add 14 H -on right sideD- Add 4 H -on right side

The correct option is A Add 14H+ on left side To balance H in Cr2O72 (aq) → Cr3+ (aq) + 7H2O (in acidic medium), we will add 14H+ on left side. Cr2O72 (aq) + ...

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Standard VII

Chemistry

Neutralisation Reaction

To balance H ...

Question

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

Add 14H⁺ on left side

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Add 4H⁺ on left side

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Add 14H⁺ on right side

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Add 4H⁺ on right side

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Solution

The correct option is A
Add 14H⁺ on left side

To balance H in Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) + 7H₂O (in acidic medium), we will add 14H⁺ on left side.

Cr₂O₇^2- _(aq) + 14H⁺ → Cr³⁺ _(aq) + 7H₂O

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Similar questions

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6Fe²⁺(aq) + Cr₂O₇^2-(aq) → 6 Fe³⁺(aq) +2Cr ³⁺(aq)

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To balance H in Cr2O72- (aq) → Cr3+ (aq) + 7H2O (in acidic medium), we will

The correct option is A Add 14H+ on left side To balance H in Cr2O72- (aq) → Cr3+ (aq) + 7H2O (in acidic medium), we will add 14H+ on left side. Cr2O72- (aq) + 14H+ → Cr3+ (aq) + 7H2O

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

The correct option is A
Add 14H⁺ on left side

To balance H in Cr₂O₇^2- _(aq) → Cr³⁺ _(aq) + 7H₂O (in acidic medium), we will add 14H⁺ on left side.

Cr₂O₇^2- _(aq) + 14H⁺ → Cr³⁺ _(aq) + 7H₂O