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nezhdanof
@nezhdanof
July 2022
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Решить ОВР в щелочной среде методом электронного баланса
K[Cr(OH)₄] + KJO₄ = /CrO₄²⁻, J⁻/
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Andreee
K[Cr(OH)₄] + KJO₄ = /CrO₄²⁻, J⁻/
8 | [Cr(OH)₄]⁻ - 3e +4OH⁻ = CrO₄²⁻ + 4H₂O
3 | JO₄⁻ + 8e +4H₂O = J⁻ + 8OH⁻
___________________________________
8[Cr(OH)₄]⁻ + 3JO₄⁻ + 8OH⁻ = 20 H₂O + 8CrO₄²⁻ + 3J⁻
8K[Cr(OH)₄] + 3KJO₄ + 8NaOH = 4K₂CrO₄ + 4Na₂CrO₄ + 3KJ + 20 H₂O
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Answers & Comments
8 | [Cr(OH)₄]⁻ - 3e +4OH⁻ = CrO₄²⁻ + 4H₂O
3 | JO₄⁻ + 8e +4H₂O = J⁻ + 8OH⁻
___________________________________
8[Cr(OH)₄]⁻ + 3JO₄⁻ + 8OH⁻ = 20 H₂O + 8CrO₄²⁻ + 3J⁻
8K[Cr(OH)₄] + 3KJO₄ + 8NaOH = 4K₂CrO₄ + 4Na₂CrO₄ + 3KJ + 20 H₂O