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1

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.
2

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

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$ = $$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$
3

AIPMT 2011 Mains

A solution contains Fe2+, Fe3+ and I$$-$$ ions. This solution was treated with iodine at 35oC. Eo for Fe3+/Fe2+ is + 0.77 V and Eo for I2/2I$$-$$ = 0.536 V.
The favourable redox reaction is
A
I2 will be reduced to I$$-$$
B
there will be no redox reaction
C
I$$-$$ will be oxidised to I2
D
Fe2+ will be oxidised to Fe3+

Explanation

Since the reduction potential of Fe3+/Fe2+ is greater than that of I2 /I , Fe3+ will be reduced and I will be oxidised.

2Fe3+ + 2I $$\to$$ 2Fe2+ + I2
4

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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