AIPMT 2010 Mains | Electrochemistry Question 46 | Chemistry

AIPMT 2010 Mains

MCQ (Single Correct Answer)

-1

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)

AIPMT 2010 Mains

MCQ (Single Correct Answer)

-1

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$$_4^{2 - }$$}

AIPMT 2010 Prelims

MCQ (Single Correct Answer)

-1

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

increase in ionic mobility of ions

100% ionisation of electrolyte at normal dilution

increase in both i.e., number of ions and ionic mobility of ions

increase in number of ions.

AIPMT 2010 Prelims

MCQ (Single Correct Answer)

-1

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25^oC. The value of standard Gibb's energy, $$\Delta $$G^o will be
(F = 96500 C mol^$$-$$1)

$$-$$ 89.0 kJ

$$-$$ 89.0 J

$$-$$ 44.5 kJ

$$-$$ 98.0 kJ

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