The value of ‘n’ in the balanced reaction:
will be;
Given:
Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O
1. Separate the Two Half Cells
Reduction:
Cr₂O₇²⁻ → Cr³⁺
Oxidation:
Fe²⁺ → Fe³⁺
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2. Balancing Atoms
Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
Fe²⁺ → Fe³⁺
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3. Balancing Charge
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Fe²⁺ → Fe³⁺ + e⁻
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4. Balancing Electrons
×6
6Fe²⁺ → 6Fe³⁺ + 6e⁻
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5. Add Both Half Cells
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
In which of the following compounds, oxygen shows positive oxidation state?
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