1
MCQ (Single Correct Answer)

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?
A
$$2\mathop \Lambda \limits^ \circ $$Al3+   +   $$3\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$
B
$$\mathop \Lambda \limits^ \circ $$Al3+   +   $$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$
C
($$\mathop \Lambda \limits^ \circ $$Al3+   +   $$\mathop \Lambda \limits^ \circ $$so$$_4^{2 - }$$) $$ \times $$ 6
D
$${1 \over 3}$$$$\mathop \Lambda \limits^ \circ $$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.
2
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.
3
MCQ (Single Correct Answer)

AIPMT 2009

Given :
(i)   Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V
(ii)  Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V
Electrode potential, Eo for the reaction,
Cu+ + e$$-$$ $$ \to $$ Cu,  will be
A
0.90 V
B
0.30 V
C
0.38 V
D
0.52 V

Explanation

For the reaction,

Cu2+ + 2e$$-$$ $$ \to $$ Cu,  Eo = 0.337 V

$$\Delta $$Go = - nFEo

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu2+ + e$$-$$ $$ \to $$ Cu+,  Eo = 0.153 V

$$\Delta $$Go = - nFEo

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu2+ + e$$-$$ $$ \to $$ Cu+

$$\Delta $$Go = –0.521 F = –nFE°

$$ \Rightarrow $$ E° = 0.52 V
4
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).
A
8.1 $$ \times $$ 104 g
B
2.4 $$ \times $$ 105 g
C
1.3 $$ \times $$ 104 g
D
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

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