Question

(iv) Consider this reaction: 3Au(s)
+8H+ + 2NO3-→ 3Auz+(aq) + 2NO
(g) + 4H2O(1)​

Answer 1

The oxidation number of fluorine in a compound is always −1. ... + 8H. + → 3Cu2+ + 2NO + 4H2O. +5 −2. +1. +2 −2 +1 −2. 2NO3. − + 6e ... H2(g) + CuO(s) → Cu(s) + H2O(l) c.

Answer 2

In this reaction:

3Au(s)+$8H^{+}$+2NO3-->3Au+ +2NO(g)+4H2O

• Au(gold) is oxidized to Au+
• N(nitrogen) is getting reduced from oxidation number +3 to +1
• Hence, it is a redox reaction where Reducing Agent is Gold and Oxidising Agent is Nitrogen.