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2017-11-06 11:07 am
Write and balance the equation for the reduction of iodate by hydrogen sulfite ions to give iodide and sulfate in basic aqueous solution.
回答 (3)
2017-11-06 11:19 am
✔ 最佳答案
Reduction half-equation : IO₃⁻(aq) + 3H₂O(l) + 6e⁻ → I⁻(aq) + 6OH⁻(aq) …… [1]Oxidation half-equation : HSO₃⁻(aq) + 3OH⁻(aq) → SO₄²⁻(aq) + 2H₂O(l) + 2e⁻ …… [2]
[1] + [2]×3, and cancel 6e⁻, 3H₂O(l) and 6OH⁻(aq) on the both sides. The balanced overall equation is :
IO₃⁻(aq) + 3HSO₃⁻(aq) + 3OH⁻(aq) → I⁻(aq) + 3SO₄²⁻(aq) + 3H₂O(l)
2017-11-06 11:22 am
IO3^- + HS^- ---> I2 + SO4^2-
half-reactions:
IO3^- ---> I2
HS^- ---> SO4^2-
Balance in acidic:
10e- + 12H^+ + 2IO3^- ---> I2 + 6H2O
4H2O + HS^- ---> SO4^2- + 9H^+ + 8e^-
Equalize electrons:
40e- + 48H^+ + 8IO3^- ---> 4I2 + 24H2O
20H2O + 5HS^- ---> 5SO4^2- + 45H^+ + 40e^-
Add:
3H^+ + 8IO3^- + 5HS^- ---> 5SO4^2- + 4I2 + 4H2O
half-reactions:
IO3^- ---> I2
HS^- ---> SO4^2-
Balance in acidic:
10e- + 12H^+ + 2IO3^- ---> I2 + 6H2O
4H2O + HS^- ---> SO4^2- + 9H^+ + 8e^-
Equalize electrons:
40e- + 48H^+ + 8IO3^- ---> 4I2 + 24H2O
20H2O + 5HS^- ---> 5SO4^2- + 45H^+ + 40e^-
Add:
3H^+ + 8IO3^- + 5HS^- ---> 5SO4^2- + 4I2 + 4H2O
2017-11-06 11:07 am
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收錄日期: 2021-04-18 17:59:43
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