(i) Balance the following equations by oxidation number method:
(1) Cu + HNO3 → Cu (NO3)2 + NO2 + H2O)
(2) K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2
(ii) Give reasons for the following:
(1) HNO3 acts only as an oxidising agent while HNO2 can act both as a
reducing agent and an oxidising agent.
(2) Chlorine liberates iodine from KI solution.
How will you convert the following? (Give balanced equation.)
(1) Sodium acetate to methane
Find the oxidation number of:
(1) S in Na2S4 O6
2) S in
Asked by defence221175 | 17th Feb, 2020, 06:10: PM
Que: HNO3 acts only as an oxidising agent while HNO2 can act both as a reducing agent and an oxidising agent.
The oxidation number of N in HNO3 is +5, which is maximum. So it can only decrease.
Therefore, HNO3 acts only as an oxidising agent.
In HNO2, the oxidation number of N is +3. It can increase by losing electrons or decrease by accepting electrons.
Therefore, HNO2 can act both as a reducing agent and an oxidising agent.
Que: Chlorine liberates iodine from KI solution.
The reaction is;
Cl2 + 2KI → 2KCl + I2
Cl2 + 2I− → 2Cl− + I2
Chlorine is strong oxidising agent than iodine. It oxidises iodide ions to iodine .
Therefore whem chlorine reacts with KI, iodine is liberated.
Answered by Varsha | 18th Feb, 2020, 04:18: PM
- oxidation State of cr2o-²7
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