12th Grade > Chemistry
REDOX REACTIONS MCQs
Total Questions : 29
| Page 3 of 3 pages
Answer: Option C. -> I2
:
C
Both KMnO4 and O2 are well known oxidizing agents. Of the given options I2 is a reducing agent.
:
C
Both KMnO4 and O2 are well known oxidizing agents. Of the given options I2 is a reducing agent.
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.
:
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.
Answer: Option C. -> A3 B4 C2 D1
:
C
Let us calculate the oxidation numbers for Nitrogen in each case:
(a) ∙NO2;x−4=0;x=+4
(b) ∙HNO;1+x−2=0;x=+1
(c) ∙NH3;x+3=0;x=−3
(d) ∙N2O5;2x−10=0;x=102;x=5
:
C
Let us calculate the oxidation numbers for Nitrogen in each case:
(a) ∙NO2;x−4=0;x=+4
(b) ∙HNO;1+x−2=0;x=+1
(c) ∙NH3;x+3=0;x=−3
(d) ∙N2O5;2x−10=0;x=102;x=5
Answer: Option C. -> Prevent action of water and salt
:
C
Magnesium provides cathodic protection and prevents rusting or corrosion thus preventing the action of water and salt.
:
C
Magnesium provides cathodic protection and prevents rusting or corrosion thus preventing the action of water and salt.
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.
:
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.
Answer: Option C. -> +6
:
C
Looking at the ion M3+ - clearly the oxidation number is +3.
Further -M3+ loses 3 electrons to be comeM6+.
Hence - the new oxidation number is 6.
:
C
Looking at the ion M3+ - clearly the oxidation number is +3.
Further -M3+ loses 3 electrons to be comeM6+.
Hence - the new oxidation number is 6.
Answer: Option A. -> H3PO3 is dibasic and reducing
:
A
H−O−H|P↓O−OH, it can donate two hydrogen ion, hence called dibasic. It is reducing as by releasing hydrogen it gets oxidised and reduceother.
:
A
H−O−H|P↓O−OH, it can donate two hydrogen ion, hence called dibasic. It is reducing as by releasing hydrogen it gets oxidised and reduceother.
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 ofCr2O2−7 = 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
:
B
First - let us write down the reaction:
Cr2O2−7+An−BD⟶Cr3++A−n+x
From stoichiometry, we have:
Equivalents ofCr2O2−7 = 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
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 theyallbecome Al3+.
So overall, there are 24 electrons transferred.
:
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 theyallbecome Al3+.
So overall, there are 24 electrons transferred.
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