NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### 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
2

### 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)
3

### 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.
4

### AIPMT 2010 Prelims

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25oC. The value of standard Gibb's energy, $$\Delta$$Go will be
(F = 96500 C mol$$-$$1)
A
$$-$$ 89.0 kJ
B
$$-$$ 89.0 J
C
$$-$$ 44.5 kJ
D
$$-$$ 98.0 kJ

## Explanation

The cell reaction

Cu + 2Ag+ $$\to$$ Cu2+ + 2Ag

We know, $$\Delta$$G° = – nFE°cell

= – 2 × 96500 × 0.46 = – 88780 J

= – 88.780 kJ = – 89 kJ

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

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

NEET

Class 12