Redox Reactions And Electrochemistry Question 698
Question: Match List I with List II and select the correct answer using the code given below the lists:
| List I | List II |
|---|---|
| Oxidising agent | (p) Disproportionation |
| $ Mn_3O_4 $ | (q) Redox reaction |
| $ C_6H_6 $ | (r) Decreases oxidation number |
| $ 2C{{u}^{+}}\xrightarrow{{}}C{{u}^{2+}} $ $ +Cu^{0} $ | (s) Fractional oxidation number |
| (E) $ H_2O_2+O_3\xrightarrow{{}} $ $ H_2O+2O_2 $ | (t) Oxidation number is -1 |
Codes:
Options:
A) A $ \to $ q, B $ \to $ p, C $ \to $ t, D $ \to $ s, E $ \to $ r
B) A $ \to $ t, B $ \to $ s, C $ \to $ r, D $ \to $ q, E $ \to $ P
C) A $ \to $ r, B $ \to $ t, C $ \to $ s, D $ \to $ p, E $ \to $ q
D) A $ \to $ r, B $ \to $ s, C $ \to $ t, D $ \to $ p, E $ \to $ q
Show Answer
Answer:
Correct Answer: D
Solution:
[d] [a] Oxidising agent, which undergo in reduction process and decreases its oxidation number. [b] It is $ MnO,Mn_2O_3 $ and O.S. $ =\frac{+8}{3} $ [c] $ C_6H_66x+6=0 $ $ x=-1 $ [d] Disproportionate reaction. (e) Redox reaction.
BETA
