Redox Reactions(11th And 12th > Chemistry ) Questions and Answers
Explanation:-
Answer: Option B. -> SO2:
B
In SO2, sulphur has a +4 oxidation state. In H2SO4, Sulphur exhibits its highest oxidation state of +6.
On the other hand - in H2S, sulphur exists in its lowest -2 oxidation state. Thus, we can surmise
that SO2 could act as both an oxidising agent as well as a reducing agent. Nitrogen exhibits its highest oxidation
state of +5 in HNO3.
Explanation:-
Answer: Option D. -> 24:
D
Elemental metallic Aluminium carries an oxidation number of 0.
From the products, we see that 0Al becomes 3+Al3+.
Clearly, Aluminium is the reducing agent.
There are 8 Aluminium atoms on the left and they all become Al3+.
So overall, there are 24 electrons transferred.
Explanation:-
Answer: Option C. -> 1, 3, 4, 5:
C
This has to be answered based on the oxidation number of the central metal atom in each case.
To start with Mn has +7 oxidation number in KMnO4. Likewise, we have
+6MnO2−4
+4MnO2
+3Mn2O3
+2Mn2+
The number of electrons transferred in each case is 1, 3, 4, 5 respectively.
Explanation:-
Answer: Option B. -> 3-n:
B
First - let us write down the reaction:
Cr2O2−7+An−BD⟶Cr3++A−n+x
From stoichiometry, we have:
Equivalents of Cr2O2−7 = equivalents of ABD
1.68×10−3=3.26×10−3×x
Solving, we get x=3.
Hence, the new oxidation number is: 3-n
Explanation:-
Answer: Option A. -> An oxidizing agent:
A
The sulphur in H2SO4 gets reduced from a +6 oxidation state to +4 in SO2.
Also, we see that Ag0 becomes Ag+1.
This shows that sulphur has gained electrons and got reduced thereby oxidizing others.
So H2SO4 acts as an oxidizing agent.