We can balance the following ionic equation by adding6 Fe 2- aq - Cr 2 O 72- aq - 6 Fe 3- aq -2 Cr 3- aq A- 3 H 2 O on LHS and 6 H -on RHSB- 7 H 2 O on LHS and 14 H -on RHS C- 3 H 2 O on RHS and 6 H -on LHSD- 7 H 2 O on RHS and 14 H -on LHS

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Standard X

Chemistry

Gravity Separation

We can balanc...

Question

We can balance the following ionic equation by adding

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

3H₂O on LHS and 6 H⁺ on RHS

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7 H₂O on LHS and 14 H⁺ on RHS

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3H₂O on RHS and 6 H⁺ on LHS

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7 H₂O on RHS and 14 H⁺ on LHS

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Solution

For balancing the ionic reaction 14 H⁺ should be added on left hand side of equation and 7H₂O on right hand side of equation.

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Q. The electrode with reaction :

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

can be represented as:

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

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We can balance the following ionic equation by adding 6Fe2+(aq) + Cr2O72-(aq) → 6 Fe3+(aq) +2Cr 3+(aq)

For balancing the ionic reaction 14 H+ should be added on left hand side of equation and 7H2O on right hand side of equation.

We can balance the following ionic equation by adding

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

For balancing the ionic reaction 14 H⁺ should be added on left hand side of equation and 7H₂O on right hand side of equation.