Topic: Redox Reactions (Test 1)



Topic: Redox Reactions
Q.1
The oxidation number of oxygen in most compounds is
A. 4
B. -3
C. 1
D. -2
Answer : Option D
Explaination / Solution:

Oxygen is the most electronegative element. Oxygen have 6 electrons in its outermost shell and to gain noble gas configuration oxygen have to take 2 more electrons.

Workspace
Report
Q.2
A metal in a compound can be displaced by another metal in the uncombined state. Which metal is a better reducing agent in such a case?
A. Both are same in reducing capacity
B. better reducing agent is the one that loses more electrons
C. better reducing agent is the one that loses less electrons
D. the reduced metal is a better reducing agent than the reducing metal
Answer : Option B
Explaination / Solution:

Reducing agent is one in which element undergo oxidation itself and make the other substance reduce.

Workspace
Report
Q.3
 represents
A. oxidation number of +2 of Zinc
B. redox couple
C. oxidised form of Zinc
D. reduced form of Zinc
Answer : Option B
Explaination / Solution:

This is redox couple.

Where following reaction take place :

Zn2+  + 2e-   Zn


Workspace
Report
Q.4
Assign oxidation number to P in 
A. +1
B. +6
C. +4
D. 5.0
Answer : Option D
Explaination / Solution:

Oxidation state of P in NaH2POis +5.

Let the oxidation number of P be x. As we already know oxidation number of other atoms so we can calculate that of P too.

Na = +1

H= +1 (+2 for hydrogen molecule)

O = -2

Now calculating oxidation number of P:-

1 +2 + x +(-8) = 0

On solving we get x= +5.


Workspace
Report
Q.5


A. hydrogen is oxidised
B. sodium is reduced
C. electronegativity of sodium determines the direction of the reaction
D. sodium is oxidised and hydrogen is reduced
Answer : Option D
Explaination / Solution:

sodium is oxidized and hydrogen is reduced.

Workspace
Report
Q.6
An oxidation number of +1 is found in all their compounds of one of the below given options
A. all alkaline earth metals
B. all alkali metals
C. all transition elements
D. superoxides
Answer : Option B
Explaination / Solution:

The alkali metals (Group 1) have 1 valence electron. Alkali metals looses this electron to achieve noble gas configuration, and so alkali metals have oxidation number +1.

Workspace
Report
Q.7

E0 values of some redox couples are given below. On the basis of these values choose the correct option. 

E0 values: Br2/Br= +1.90; Ag+/Ag = +0.80; Cu2+/Cu = +0.34; I2/I-(s) = +0.54


A. Cu will reduce 
B. Cu will reduce Ag
C. Cu will reduce Br2
D. Cu will reduce 
Answer : Option C
Explaination / Solution:

Because in electrochemical series Cu is above all in sequence I2, Ag and Br2, therefore, Cu will reduce Br2.

Workspace
Report
Q.8
Consider the elements: Cs, Ne, I and F. Identify the element which exhibits neither the negative nor does the positive oxidation state.
A. Cs
B. Ne
C. Ne and F
D. F
Answer : Option B
Explaination / Solution:

Ne is a noble gas, so does not show any positive or negative oxidation state.

Workspace
Report
Q.9
The electron releasing tendency of the metals, zinc, copper and silver is in the order:
A. Zn>Cu>Ag
B. Cu>Ag>Zn
C. Zn>Ag>Cu
D. Ag>Cu>Zn
Answer : Option A
Explaination / Solution:

The electron releasing tendency of the metals, zinc, copper and silver is in the order Zn>Cu>Ag This is according to the metal activity series or electrochemical series.

Workspace
Report
Q.10
Copper nitrate is a blue coloured solution. Place a strip of metallic zinc in an aqueous solution of copper nitrate for about one hour. What happens?
A. The aqueous solution of copper nitrate turns green in colour and mettallic zinc strip turns darker in colour.
B. zinc strip becomes coated with blue colour
C. the blue colour of the solution turns to red
D. the blue colour of the solution becomes more intense blue.
Answer : Option A
Explaination / Solution:

On placing a strip of mettallic zinc in an aqueous solution of copper nitrate for about one hour, the copper nitrate solution turns green in colour and zinc strip turns darker. 

Formula and Ionic equation for the reaction is as follows.

Formula Equation:Zn (s) + Cu(NO3)2 (aq) → Cu (s) + Zn(NO3)2 (aq)
Ionic Equation:Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
 

As is seen in ionic equation Zn is loosing 2 electrons, thus oxidation  occurs  and Cu is gaining 2 electrons, thus reduction occurs. Since oxidation and reduction takes place simultaneously in this reaction, it is a redox reaction. Also since Zn is loosing electrons and becoming darker in colour, it is the reducing agent and Cu is gaining 2 electrons, it is the reducing agent. The resultant solution Zinc nitrate [Zn(NO3)2] is green in colour.


Workspace
Report