1
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

The electrode potentials for Cu2+(aq) + e$$-$$ $$ \to $$ Cu+(aq)
and Cu+(aq) + e$$-$$ $$ \to $$ Cu(s) are + 0.15 V and + 0.50 V respectively.

The value of Eocu2+/cu will be
A
0.500 V
B
0.325 V
C
0.650 V
D
0.150 V

Explanation

Cu2+(aq) + e$$-$$ $$ \to $$ Cu+(aq) ; E1o = 0.15 V

Cu+(aq) + e$$-$$ $$ \to $$ Cu(s) ; E2o = 0.50 V

Cu2+ + 2e $$ \to $$ Cu ; Eo = ?

$$\Delta $$Go = $$\Delta $$G1° + $$\Delta $$G2°

$$ \Rightarrow $$ – nFE° = – n1FE1° – n2FE2°

$$ \Rightarrow $$ Eo = $${{1 \times 0.15 + 1 \times 0.50} \over 2}$$ = 0.325 V
2
MCQ (Single Correct Answer)

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

Explanation

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
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Standard electrode potential of three metals X, Y and Z are $$-$$1.2 V, + 0.5 V and $$-$$ 3.0 V respectively. The reducing power of these metals will be
A
Y > Z > X
B
Y > X > Z
C
Z > X > Y
D
X > Y > Z

Explanation

As the electrode potential drops, reducing power increases.

So, Z (–3.0 V) > X (–1.2 V) > Y (+ 0.5 V)
4
MCQ (Single Correct Answer)

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)

Explanation

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.

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI