Using the redox reaction below determine which element is oxidized and which is reduced. 4nh3 + 3ca (clo) 2 → 2n2 + 6h2o + 3cacl2 n is oxidized and cl is reduced cl is oxidized and o is reduced cl is oxidized and n is reduced h is oxidized and n is reduced n is oxidized and o is reduced

Question

Jaxson Blackwell

Chemistry

11 February, 04:05

Using the redox reaction below determine which element is oxidized and which is reduced. 4nh3 + 3ca (clo) 2 → 2n2 + 6h2o + 3cacl2 n is oxidized and cl is reduced cl is oxidized and o is reduced cl is oxidized and n is reduced h is oxidized and n is reduced n is oxidized and o is reduced

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Deon Buckley · Answer 1

The oxidation number of elements in equation below are,

4NH₃ + 3Ca (ClO) ₂ → 2N₂ + 6H₂O + 3CaCl₂

O. N of N in NH₃ = - 3

O. N of Ca in Ca (ClO) ₂ and CaCl₂ = + 2

O. N of N in N₂ = 0

O. N of Cl in Ca (ClO) ₂ = + 1

O. N of Cl in CaCl₂ = - 1

Oxidation:

Oxidation number of Nitrogen is increasing from - 3 (NH₃) to 0 (N₂).

Reduction:

Oxidation number of Cl is decreasing from + 1 [Ca (ClO) ₂] to - 1 (CaCl₂).

Result:

N is oxidized and Cl is reduced.