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QUESTION: 1

The equilibrium constant of the reaction:

E° = 0.46 V at 298 K is [2007]

Solution:

or K_{c }= Antilog 15.57 = 3.7 × 10^{15} ≈ 4 × 10^{15}

QUESTION: 2

On the basis of the following E° values, the strongest oxidizing agent is : [2008]

Solution:

For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of Fe^{+2} is less than that of [Fe(CN)_{6}]^{4−}. Therefore, Fe^{+2} is stronger oxidizing agent than [Fe(CN)_{6}]^{4−}. Also, the stronger oxidizing agent should easily reduce itself. Here, Fe^{+3} is easily reduced than Fe^{+2}. Therefore, among all the four, Fe^{+3} is the stronger oxidizing agent.

QUESTION: 3

Kohlrausch’s law states that at : [2008]

Solution:

Kohlrausch ’s Law states that at infinite dilution, each ion migrates independently of its co-ion and contributes to the total equivalent conductance of an eletrolyte a definite share which depends only on its own nature.

From this definition we can see that option (d) is the correct answer.

QUESTION: 4

Standard free energies of formation (in kJ/mol) at 298 K are – 237.2, – 394.4 and – 8.2 for H_{2}O(l), CO_{2}(g) and pentane (g), respectively. The value E°cell for the pentane-oxygen fuel cell is : [2008]

Solution:

Writing the equation for pentane-oxygen fuel cell at respective electrodes and over all reaction, we get At Anode:

At Cathode:

Calculation of ΔG° for the above reaction

ΔG° = [5×(–394.4) + 6× (–237.2)]

– [–8.2]

= – 1972.0 – 1423.2 + 8.2 = – 3387.0 kJ

= – 3387000 Joules.

From the equation we find n = 32

Using the relation,

and substituting various values, we get

Thus option (c) is correct answer.

QUESTION: 5

Given: [2009]

Electrode potential, Eo for the reaction, will be :

Solution:

= – 2 × F × 0.337

= – 0.674 F ....(i)

= – 1 × F × – 0.153

= 0.153 F ....(ii)

On adding eqn (i) & (ii)

Here n = 1 Δ E^{o} = + 0.52 V

QUESTION: 6

Al 2O_{3} is reduced by electrolysisat low potentials and high currents. If 4.0 × 10^{4} amperes of current is passed through molten Al2O_{3} for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol^{–1}) [2009]

Solution:

Now since 96500 C liberates 9 g of Al

8.64 × 10^{8} C liberates

= 8.1 × 10^{4 }g of Al

QUESTION: 7

The equivalent conductance of solution ofa weak monobasic acid is 8.0 mhos cm^{2} and at infinite dilution is 400 mhos cm^{2}. The dissociation constant of this acid is: [2009]

Solution:

Degree of dissociation

= 1.25 x 10^{-5}

QUESTION: 8

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V at 25°C. The value of standard Gibbs energy, ΔG^{0} will be (F = 96500 C mol ^{–1})

Solution:

Here, n = 2 ,

QUESTION: 9

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to: [2010]

Solution:

Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions which in turn increases equivalent conductance of the solution.

QUESTION: 10

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al_{2} (SO_{4})_{3}, Given that and are the equivalent conductances at infinite dilution of the respective ions? [2010]

Solution:

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.

∴ Correct choice : (c)

QUESTION: 11

Consider the following relations for emf of a electrochemical cell: [2010]

(i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)

(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)

(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)

(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)

Which of the above relations are correct?

Solution:

Option (b) and (d) are correct

∴ Correct choice : (a)

QUESTION: 12

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 : [2011]

Solution:

As the value of standard reduction potential decreases the reducing power increases i.e.,

QUESTION: 13

The electrode potentials for [2011]

are + 0.15 V and + 0.50, respectively. The value of E° _{Cu 2+ / Cu }will be :

Solution:

⇒ 0.325

QUESTION: 14

Standard electrode potential for Sn^{4+} / Sn^{2+} couple is + 0.15 V and that for the Cr^{3+} / Cr couple is – 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be : [2011]

Solution:

= 0.15 – (– 0.74)

= + 0.89 V

QUESTION: 15

If the E°_{cell} for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG° and K_{eq} ?[2011]

Solution:

Standard Gibbs free energy is given as ΔG°

QUESTION: 16

A solution contains Fe^{2+}, Fe^{3+} and I^{– }ions. This solution was treated with iodine at 35°C. E° for Fe^{3+} / Fe^{2+} is + 0.77 V and E° for I_{2}/2I^{–} = 0.536 V.The favourable redox reaction is : [2011 M]

Solution:

= 0.77 – 0.536

= 0.164 V

So, reaction will taken place.

QUESTION: 17

Limiting molar conductivity of NH_{4}OH

is equal to : [2012]

Solution:

QUESTION: 18

Molar conductivities at infinite dilution of NaCl, HCl and CH_{3}COONa are 126.4, 425.9 and 91.0 S cm^{2} mol^{–1} respectively. for CH_{3}COOH will be : [2012 M]

Solution:

= 91 + 425.9 – 126.4 = 390.5

QUESTION: 19

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ? [NEET 2013]

Solution:

QUESTION: 20

At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm^{-1} cm^{2}mol^{-1} and at infinite dilution its molar conductance is 238 ohm^{-1} cm^{2} mol^{-1}. The degree or ionisation of ammonium hydroxide at the same concentration and temperature is :[NEET 2013]

Solution:

QUESTION: 21

A button cell used in watches function s as following

If half cell potentials are :

The cell potential will be : [NEET 2013]

Solution:

E°_{Cell} = E°_{OP }+ E°_{RP} = 0.76 + 0.314 = 1.10 V

QUESTION: 22

How many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1 Faraday = 96,500 C; Atomic mass of Co = 59 u) [NEET Kar. 2013]

Solution:

Applying,

Equivalent weight of cobalt (II) = 59/2

I = 10 A

Time (t) = 109 min = 109 × 60 sec

Substituting these values we get,

QUESTION: 23

Consider the half-cell reduction reaction :

The E° for the reaction and possibility of the forward reaction are, respectively [NEET Kar. 2013]

Solution:

ΔE° = E°_{red} + E°_{oxd} = – 1.81 – 1.51

= – 2.69

Since ΔE° is negative

∴ ΔG = –nFE°, ΔG will have positive value so, forward reaction is not possible.

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