A) Evolved \[{{I}_{2}}\] is reduced
B) \[C{{u}_{2}}{{I}_{2}}\] is formed
C) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]is oxidized
D) \[Cu{{I}_{2}}\]is formed
Correct Answer: B
Solution :
[b] \[2CuS{{O}_{4}}+4KI\to 2{{K}_{2}}S{{O}_{4}}+C{{u}_{2}}{{I}_{2}}+{{I}_{2}}\] \[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\]
You need to login to perform this action.
You will be redirected in
3 sec
