1
MCQ (Single Correct Answer)

AIPMT 2012 Mains

The Gibb's energy for the decomposition of Al2O3 at 500oC is as follows
$${2 \over 3}$$ Al2O3 $$ \to $$ $${4 \over 3}$$ Al + O2
$$\Delta $$rG = +960 kJ mol$$-$$1
The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500oC is at least
A
4.5 V
B
3.0 V
C
2.5 V
D
5.0 V

Explanation

We know,

$$\Delta $$Go = – nFEo

$${2 \over 3}$$ Al2O3 $$ \to $$ $${4 \over 3}$$ Al + O2

Total number of Al atoms in Al2O3

= $${2 \over 3} \times 2 = {4 \over 3}$$

Al3+ + 3e $$ \to $$ Al

As 3e change occur for each Al atom

$$ \therefore $$ n = $${4 \over 3} \times 3 = 4$$

Eo = - $${{\Delta G^\circ } \over {nF}}$$

= - $${{960 \times 1000} \over {4 \times 96500}}$$

= - 2.5 V
2
MCQ (Single Correct Answer)

AIPMT 2012 Mains

Standard reduction potentials of the half reactions are given below :
F2(g) + 2e$$-$$ $$ \to $$ 2F$$-$$(aq) ;   Eo = + 2.85 V
Cl2(g) + 2e$$-$$ $$ \to $$ 2Cl$$-$$(aq) ;   Eo = + 1.36 V
Br2(l) + 2e$$-$$ $$ \to $$ 2Br$$-$$(aq) ;   Eo = + 1.06 V
I2(s) + 2e$$-$$ $$ \to $$ 2I$$-$$(aq) ;  Eo = + 0.53 V

The strongest oxidising and reducing agents 23 respectively are
A
F2 and I$$-$$
B
Br2 and Cl$$-$$
C
Cl2 and Br$$-$$
D
Cl2 and I2

Explanation

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.
3
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

Limiting molar conductivity of NH4OH
$$\left[ {} \right.$$i.e.  $$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$$$\left. {} \right]$$ is equal to
A
$$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0$$
B
$$\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0$$
C
$$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0$$
D
$$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$

Explanation

According to Kohlrausch’s law, the molar conductivity of NH4OH

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$ = $$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$
4
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

The electrode potentials for Cu2+(aq) + e$$-$$ $$ \to $$ Cu+(aq)
and Cu+(aq) + e$$-$$ $$ \to $$ Cu(s) are + 0.15 V and + 0.50 V respectively.

The value of Eocu2+/cu will be
A
0.500 V
B
0.325 V
C
0.650 V
D
0.150 V

Explanation

Cu2+(aq) + e$$-$$ $$ \to $$ Cu+(aq) ; E1o = 0.15 V

Cu+(aq) + e$$-$$ $$ \to $$ Cu(s) ; E2o = 0.50 V

Cu2+ + 2e $$ \to $$ Cu ; Eo = ?

$$\Delta $$Go = $$\Delta $$G1° + $$\Delta $$G2°

$$ \Rightarrow $$ – nFE° = – n1FE1° – n2FE2°

$$ \Rightarrow $$ Eo = $${{1 \times 0.15 + 1 \times 0.50} \over 2}$$ = 0.325 V

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