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
2
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), CO2(g) and pentane (g) respectively. The value of Eocell for the pentane-oxygen fuel cell is
Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
B
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C
Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D
Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.
Explanation
Kohlrausch’s law states that the equivalent
conductance of an electrolyte at infinite dilution is
equal to the sum of the equivalent conductance of
the component ions.