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.
2
MCQ (Single Correct Answer)
AIPMT 2010 Prelims
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
A
increase in ionic mobility of ions
B
100% ionisation of electrolyte at normal dilution
C
increase in both i.e., number of ions and ionic mobility of ions
D
increase in number of ions.
Explanation
Dilution of strong electrolytes increases
ionisation, hence ionic mobility of ions
increases which in turn increases equivalent
conductance of the solution.
3
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?
Equivalent conductance of an electrolyte
at infinite dilution is given by the sum of
equivalent conductances of the respective ions
at infinite dilution.
4
MCQ (Single Correct Answer)
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
Questions Asked from Electrochemistry
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