AIPMT 2010 Mains

Which of the following expressions correctly represents the equivalent conductance at infinite diluation of Al₂(SO₄)₃. Given that $$\mathop \Lambda \limits^ \circ $$_Al³⁺ and $$\mathop \Lambda \limits^ \circ $$_{so$$_4^{2 - }$$} are the equivalent conductances at infinite dilution of the respective ions?

$$2\mathop \Lambda \limits^ \circ $$_Al³⁺ + $$3\mathop \Lambda \limits^ \circ $$_{so$$_4^{2 - }$$}

$$\mathop \Lambda \limits^ \circ $$_Al³⁺ + $$\mathop \Lambda \limits^ \circ $$_{so$$_4^{2 - }$$}

($$\mathop \Lambda \limits^ \circ $$_Al³⁺ + $$\mathop \Lambda \limits^ \circ $$_{so$$_4^{2 - }$$}) $$ \times $$ 6

$${1 \over 3}$$$$\mathop \Lambda \limits^ \circ $$_Al³⁺ + $${1 \over 2}$$$$\mathop \Lambda \limits^ \circ $$_{so$$_4^{2 - }$$}

Explanation

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.

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?

(iii) and (i)

(i) and (ii)

(iii) and (iv)

(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.

MCQ (Single Correct Answer)

AIPMT 2009

Given :
(i)   Cu²⁺ + 2e^$$-$$ $$ \to $$ Cu,  E^o = 0.337 V
(ii)  Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺,  E^o = 0.153 V
Electrode potential, E^o for the reaction,
Cu⁺ + e^$$-$$ $$ \to $$ Cu,  will be

0.90 V

0.30 V

0.38 V

0.52 V

Explanation

For the reaction,

Cu²⁺ + 2e^$$-$$ $$ \to $$ Cu, E^o = 0.337 V

$$\Delta $$G^o = - nFE^o

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺, E^o = 0.153 V

$$\Delta $$G^o = - nFE^o

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu²⁺ + e^$$-$$ $$ \to $$ Cu⁺

$$\Delta $$G^o = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V

MCQ (Single Correct Answer)

AIPMT 2009

Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 $$ \times $$ 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is product? (Assume 100% current efficiency, at mass of Al = 27 g mol^$$-$$1).

8.1 $$ \times $$ 10⁴ g

2.4 $$ \times $$ 10⁵ g

1.3 $$ \times $$ 10⁴ g

9.0 $$ \times $$ 10³ g

Explanation

E = Z × 96500

$$ \Rightarrow $$ $${{27} \over 3}$$ = Z $$ \times $$ 96500

$$ \Rightarrow $$ Z = $${9 \over {96500}}$$

Now applying the formula, W = Z × I × t

W = $${9 \over {96500}}$$ $$ \times $$ 4 $$ \times $$ 10⁴ $$ \times $$ 6 $$ \times $$ 60 $$ \times $$ 60

= 8.1 × 10⁴ g