Home Courses Contact About


Topic 1 : Galvanic Cell 2

9. ⇒ (AIPMT 2011 Prelims )

Standard electrode potential for Sn4+/Sn2+ couple is + 0.15 V and that for the Cr3+/Cr couple is 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be

(A) + 1.19 V

(B) + 0.89 V

(C) + 0.18 V

(D) + 1.83 V

Correct Answer is Option (B)

Sn4+/Sn2+ = 0.15 V

Cr3+/Cr = –0.74 V

cell = E°cathode – E°anode

= 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V

10. ⇒ (AIPMT 2010 Mains )

Consider the following relations for emf of an electrochemical cell
(i)   EMF of cell = (Oxidation potential of anode) (Reduction potential of cathode)
(ii)  EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) (Oxidation potential of cathode)

Which of the above relations are correct?

(A) (iii) and (i)

(B) (i) and (ii)

(C) (iii) and (iv)

(D) (ii) and (iv)

Correct Answer is Option (D)

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode – Oxidation potential of cathode.

11. ⇒ (AIPMT 2008 )

On the basis of the following Eo values, the strongest oxidizing agent is
[Fe(CN)6]4 [Fe(CN)6]3 + e ;  Eo = 0.35 V

Fe2+ Fe3+ + e ;  Eo = 0.77 V

(A) Fe3+

(B) [Fe(CN)6]3

(C) [Fe(CN)6]4

(D) Fe2+

Correct Answer is Option (A)

Substances which have higher reduction potential are stronger oxidizing agent.

[Fe(CN)6]4 [Fe(CN)6]3 + e ;  Eo = 0.35 V

Fe2+ Fe3+ + e ;  Eo = 0.77 V

Higher the +ve reduction potential, stronger will be the oxidising agent. Oxidising agent oxidises other compounds and gets itself reduced easily.

12. ⇒ (AIPMT 2006 )

EoFe2+/Fe = 0.441 V and EoFe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe3+ 3Fe2+ will be

(A) 0.111 V

(B) 0.330 V

(C) 1.653 V

(D) 1.212 V

Correct Answer is Option (D)

At anode : Fe Fe2+ + 2e-; Eo = -0.441 V

At cathode : Fe3+ + e Fe2+; Eo = 0.771 V

Fe + 2Fe3+ 3Fe2+; Eo = ?

To get the above equation, (ii) × 2 – (i)

2Fe3+ + 2e 2Fe2+; Eo = 0.771 V

Fe Fe2+ + 2e-; Eo = - 0.441 V
------------------------------------------------
Fe + 2Fe3+ 3Fe2+

Eo = 0.771 + 0.441 = 1.212 V

13. ⇒ (AIPMT 2006 )

A hypothetical electrochemical cell is shown below.

A | A + ( x M ) | | B + ( y M ) | B

The emf measured is + 0.20 V. The cell reaction is

(A) A + B+ A+ + B

(B) A+ + B A + B+

(C) A+ + e A;  B+ + e B

(D) the cell reaction cannot be predicted.

Correct Answer is Option (A)

From the given expression:

At anode : A A+ + e

At cathode : B+ + e B

Overall reaction is : A + B+ A+ + B

14. ⇒ (AIPMT 2001 )

Standard electrode potentials are
Fe2+/Fe [Eo = 0.44] and
Fe3+/Fe2+[ Eo = 0.77];
If Fe2+, Fe3+ and Fe blocks are kept together, then

(A) Fe3+ increases

(B) Fe3+ decreases

(C) Fe2+/Fe3+ remains unchanged

(D) Fe2+ decreases.

Correct Answer is Option (B)

The metals having higher negative values of their electrode potential can displace metals having lower values from their salt solutions.