How many $H_{2} O$ should be added to the RHS of the following equation to balance number of oxygen atoms on the left ?

$6 F e^{2 +} + C r_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 6 F e^{3 +} + 2 C r^{3 +} +$ ___________

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Solution

The correct option is D
7

As you can check in the equation at the reactant side,

No of O = 7

$∴$ Number of $H_{2} O$ to be added =7

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Similar questions

How many $H_{2} O$ should be added to the RHS of the following equation to balance number of oxygen atoms on the left ?

$6 F e^{2 +} + C r_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 6 F e^{3 +} + 2 C r^{3 +} +$ ___________

We can balance the following ionic equation by adding

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

Q. Given a reaction:

{C r}_{2} O_{7}^{2 -} + {F e}^{2 +} + H^{\oplus} ⟶ {C r}^{3 +} + {F e}^{3 +} + H_{2} O

The balanced equation is

6 {F e}^{2 +} + 14 H^{\oplus} + {C r}_{2} O_{7}^{2 -} ⟶ 6 {F e}^{3 +} + 2 {C r}^{3 +} + 7 H_{2} O

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If true enter 1, else enter 0.

Write half-reactions using electrons.

{C r}_{2} O_{2}^{7 -} + 6 {F e}^{2 +} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 6 {F e}^{3 +} + 7 H_{2} O

Q. The standard reaction, Gibbs energy for the electrochemical reaction.

C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O i s - 793 k J m o l^{- 1}

The standard cell potential is:

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How many H2O should be added to the RHS of the following equation to balance number of oxygen atoms on the left ? 6Fe2++Cr2O2−7+14H+⟶6Fe3++2Cr3++ ___________

The correct option is D 7 As you can check in the equation at the reactant side, No of O = 7 ∴ Number of H2O to be added =7

How many $H_{2} O$ should be added to the RHS of the following equation to balance number of oxygen atoms on the left ?

$6 F e^{2 +} + C r_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 6 F e^{3 +} + 2 C r^{3 +} +$ ___________

The correct option is D
7

As you can check in the equation at the reactant side,

No of O = 7

$∴$ Number of $H_{2} O$ to be added =7