NCERT Solutions for Class 12 Chemistry Chapter 3 –
Electrochemistry
Q 3.1:
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al,
Cu, Fe, Mg and Zn.
Answer:
According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn,
Fe, and Cu.
Mg: Al: Zn: Fe: Cu
Q 3.2:
Given the standard electrode potentials.
K+/K = –2.93V
Ag+/Ag = 0.80V
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V
Cr3+/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.
Ans:
The reducing power increases with the lowering of the reduction potential. In order of given standard electrode
potential (increasing order): K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag
Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K
Q 3.3 :
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) →Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans :
The galvanic cell in which the given reaction takes place is depicted as
(i) The negatively charged electrode is the Zn electrode (anode).
(ii) The current carriers in the cell are ions. Current flows to zinc from silver in the external circuit.
, NCERT Solutions for Class 12 Chemistry Chapter 3 –
Electrochemistry
(iii) Reaction at the anode is given by
Reaction at the anode is given by
Q 3.4:
Calculate the standard cell potentials of the galvanic cell in which the following
reactions take place.
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆rGJ and equilibrium constant of the reactions.
Ans :
(i)
= -0.40 V
The galvanic cell of the given reaction is depicted as
Now, the standard cell potential is
= – 0.40 – ( -0.74 )
= + 0.34 V
In the given equation, n = 6
F = 96487 C mol−1
, NCERT Solutions for Class 12 Chemistry Chapter 3 –
Electrochemistry
= + 0.34 V
Then,
= −6 × 96487 C mol−1 × 0.34 V
= −196833.48 CV mol−1
= −196833.48 J mol−1
= −196.83 kJ mol−1
Again,
= 34.496
K = antilog (34.496) = 3.13 × 1034
The galvanic cell of the given reaction is depicted as
Now, the standard cell potential is
Here, n = 1
Then,
= −1 × 96487 C mol−1 × 0.03 V
= −2894.61 J mol−1
Electrochemistry
Q 3.1:
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al,
Cu, Fe, Mg and Zn.
Answer:
According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn,
Fe, and Cu.
Mg: Al: Zn: Fe: Cu
Q 3.2:
Given the standard electrode potentials.
K+/K = –2.93V
Ag+/Ag = 0.80V
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V
Cr3+/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.
Ans:
The reducing power increases with the lowering of the reduction potential. In order of given standard electrode
potential (increasing order): K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag
Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K
Q 3.3 :
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) →Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans :
The galvanic cell in which the given reaction takes place is depicted as
(i) The negatively charged electrode is the Zn electrode (anode).
(ii) The current carriers in the cell are ions. Current flows to zinc from silver in the external circuit.
, NCERT Solutions for Class 12 Chemistry Chapter 3 –
Electrochemistry
(iii) Reaction at the anode is given by
Reaction at the anode is given by
Q 3.4:
Calculate the standard cell potentials of the galvanic cell in which the following
reactions take place.
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆rGJ and equilibrium constant of the reactions.
Ans :
(i)
= -0.40 V
The galvanic cell of the given reaction is depicted as
Now, the standard cell potential is
= – 0.40 – ( -0.74 )
= + 0.34 V
In the given equation, n = 6
F = 96487 C mol−1
, NCERT Solutions for Class 12 Chemistry Chapter 3 –
Electrochemistry
= + 0.34 V
Then,
= −6 × 96487 C mol−1 × 0.34 V
= −196833.48 CV mol−1
= −196833.48 J mol−1
= −196.83 kJ mol−1
Again,
= 34.496
K = antilog (34.496) = 3.13 × 1034
The galvanic cell of the given reaction is depicted as
Now, the standard cell potential is
Here, n = 1
Then,
= −1 × 96487 C mol−1 × 0.03 V
= −2894.61 J mol−1
