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Hello... :)▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃please help me in the above attachment.. No unnecessary answers❗️▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃ |
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Answer» solution: So₂ + Fe⁺³ → So₄²⁻ + Fe²⁺Here, O.N of So² is increasing, so, their is oxidation of So₂ similarly, O.N of Fe⁺³ is decreasing. so, their is Reduction of Fe⁺³OXIDATION HALF: So₂ → So₄²⁻ => So₂ + 2H₂O → So₄²⁻ [Balancing Oxygen] => So₂ + 2H₂O → So₄²⁻ + 4H⁺ [Balancing HYDROGEN] => So₂ + 2H₂O → So₄²⁻ + 4H⁺ + 2e⁻ [Balancing charge]...(1) REDUCTION HALF : Fe³⁺ → Fe²⁺ => Fe³⁺ + e⁻ → Fe²⁺ [Balancing charge] => 2Fe³⁺ + 2e⁻ → 2Fe²⁺ [Multiplying by 2]...(1) ADDING 1 & 2 So₂ + 2H₂0 + 2Fe⁺³ + 2e⁻ → So₄²⁻ + 4H⁺ + 2e⁻ + 2Fe²⁺ So₂ + 2H₂0 + 2Fe⁺³ → So₄²⁻ + 4H⁺ + 2Fe²⁺ [All things GET balanced] ________________ Amrit____ |
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