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1

### 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

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$ = $$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$
2

### 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
3

### 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
4

### AIPMT 2011 Prelims

Standard electrode potential for Sn4+/Sn2+ couple is + 0.15 V and that for the Cr3+/Cr couple is $$-$$ 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
A
+ 1.19 V
B
+ 0.89 V
C
+ 0.18 V
D
+ 1.83 V

## Explanation

Sn4+/Sn2+ = 0.15 V

Cr3+/Cr = –0.74 V

cell = E°cathode – E°anode

= 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V

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