AIPMT 2009

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MCQ (Single Correct Answer)

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

AIPMT 2009

MCQ (Single Correct Answer)

The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm² and at infinite dilution is 400 mho cm². The dissociation constant of this acid is

1.25 $$ \times $$ 10^$$-$$6

6.25 $$ \times $$ 10^$$-$$4

1.25 $$ \times $$ 10^$$-$$4

1.25 $$ \times $$ 10^$$-$$5

Explanation

The degree of dissociation ($$\alpha $$)

$$\alpha $$ = $${8 \over {400}}$$ = 2 $$ \times $$ 10^-2

From Ostwald’s dilution law for weak monobasic acid, we have

k_c = C$$\alpha $$²

= $${1 \over {32}} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2}$$

= 1.25 $$ \times $$ 10^-5

AIPMT 2009

MCQ (Single Correct Answer)

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

AIPMT 2008

MCQ (Single Correct Answer)

Standard free energies of formation (in kJ/mol) at 298 K are $$-$$237.2, $$-$$ 394.4 and $$-$$8.2 for H₂O_(l), CO_2(g) and pentane _(g) respectively. The value of E^o_cell for the pentane-oxygen fuel cell is

1.0968 V

0.0968 V

1.968 V

2.0968 V

Explanation

At Anode:

C₅H₁₂ + 10H₂O $$ \to $$ 5CO₂ + 32H⁺ + 32e^-

At Cathode:

8O₂ + 32H⁺ + 32e^- $$ \to $$ 16H₂O
-------------------------------------------------
C₅H₁₂(g) + 8O₂(g) $$ \to $$ 5CO₂(g) + 6H₂O(l)

$$\Delta $$G = 5×$$\Delta $$G_CO₂ + 6 $$\Delta $$G_(H₂O) – [$$\Delta $$G_{(C₅H₁₂)} +8 × $$\Delta $$G_O₂ ]

= 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

= – 3387 kJ mol^–1

= – 3387 × 10³ J mol^–1

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

From the overall equation we find n = 32

– 3387 × 10³ = -32 $$ \times $$ 96500 $$ \times $$ E^o_cell

E^o_cell = $${{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}$$ = 1.0968 V