AIPMT 2012 Prelims | Electrochemistry Question 41 | Chemistry

AIPMT 2012 Prelims

MCQ (Single Correct Answer)

-1

Limiting molar conductivity of NH₄OH
$$\left[ {} \right.$$i.e. $$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$$$\left. {} \right]$$ is equal to

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

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

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0$$

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

AIPMT 2011 Mains

MCQ (Single Correct Answer)

-1

A solution contains Fe²⁺, Fe³⁺ and I^$$-$$ ions. This solution was treated with iodine at 35^oC. E^o for Fe³⁺/Fe²⁺ is + 0.77 V and E^o for I₂/2I^$$-$$ = 0.536 V.
The favourable redox reaction is

I₂ will be reduced to I^$$-$$

there will be no redox reaction

I^$$-$$ will be oxidised to I₂

Fe²⁺ will be oxidised to Fe³⁺

AIPMT 2011 Prelims

MCQ (Single Correct Answer)

-1

Standard electrode potential of three metals X, Y and Z are $$-$$1.2 V, + 0.5 V and $$-$$ 3.0 V respectively. The reducing power of these metals will be

Y > Z > X

Y > X > Z

Z > X > Y

X > Y > Z

AIPMT 2011 Prelims

MCQ (Single Correct Answer)

-1

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

+ 1.19 V

+ 0.89 V

+ 0.18 V

+ 1.83 V

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