In the above reaction, the oxidation state of chlorine changes from - to -
How many electrons are transferred in the reaction?
3Cl2 + 10CrO42-+ 44H+10Cr3+ + 6ClO3-+ 22H2O?
2016-05-31 12:41 am
回答 (1)
2016-05-31 12:50 am
3Cl₂ + 10CrO₄²⁻ + 44H⁺ → 10Cr³⁺ + 6ClO₃⁻+ 22H₂O
Oxidation state of Cl in Cl₂ = 0
Oxidation state of Cl in ClO₃⁻ = +5
The oxidation state of Cl change for 0 to +5.
Number of Cl atom involved in the change = 6
Number of electrons transferred = [(+5) - 0] × 6 = 30
OR:
Oxidation state of Cr in CrO₄²⁻ = +6
Oxidation state of Cr in Cr³⁺ = +3
The oxidation state of Cl change for +6 to +3.
Number of Cl atom involved in the change = 10
Number of electrons transferred = [(+6) - (+3)] × 10 = 30
Oxidation state of Cl in Cl₂ = 0
Oxidation state of Cl in ClO₃⁻ = +5
The oxidation state of Cl change for 0 to +5.
Number of Cl atom involved in the change = 6
Number of electrons transferred = [(+5) - 0] × 6 = 30
OR:
Oxidation state of Cr in CrO₄²⁻ = +6
Oxidation state of Cr in Cr³⁺ = +3
The oxidation state of Cl change for +6 to +3.
Number of Cl atom involved in the change = 10
Number of electrons transferred = [(+6) - (+3)] × 10 = 30
收錄日期: 2021-04-18 14:55:20
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