The equilibrium constant of the reaction:
E° = 0.46 V at 298 K is 
or Kc = Antilog 15.57 = 3.7 × 1015 ≈ 4 × 1015
On the basis of the following E° values, the strongest oxidizing agent is : 
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.
Kohlrausch’s law states that at : 
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.
Standard free energies of formation (in kJ/mol) at 298 K are – 237.2, – 394.4 and – 8.2 for H2O(l), CO2(g) and pentane (g), respectively. The value E°cell for the pentane-oxygen fuel cell is : 
Writing the equation for pentane-oxygen fuel cell at respective electrodes and over all reaction, we get At Anode:
Calculation of ΔG° for the above reaction
ΔG° = [5×(–394.4) + 6× (–237.2)]
= – 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.
Electrode potential, Eo for the reaction, will be :
= – 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 Eo = + 0.52 V
Al 2O3 is reduced by electrolysisat low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol–1) 
Now since 96500 C liberates 9 g of Al
8.64 × 108 C liberates
= 8.1 × 104 g of Al
The equivalent conductance of solution ofa weak monobasic acid is 8.0 mhos cm2 and at infinite dilution is 400 mhos cm2. The dissociation constant of this acid is: 
Degree of dissociation
= 1.25 x 10-5
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, ΔG0 will be (F = 96500 C mol –1)
Here, n = 2 ,
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to: 
Dilution of strong electrolytes increases ionisation, hence ionic mobility of ions which in turn increases equivalent conductance of the solution.
Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2 (SO4)3, Given that and are the equivalent conductances at infinite dilution of the respective ions? 
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)
Consider the following relations for emf of a electrochemical cell: 
(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?
Option (b) and (d) are correct
∴ Correct choice : (a)
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 : 
As the value of standard reduction potential decreases the reducing power increases i.e.,
The electrode potentials for 
are + 0.15 V and + 0.50, respectively. The value of E° Cu 2+ / Cu will be :
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 : 
= 0.15 – (– 0.74)
= + 0.89 V
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 Keq ?
Standard Gibbs free energy is given as ΔG°
A solution contains Fe2+, Fe3+ and I– ions. This solution was treated with iodine at 35°C. E° for Fe3+ / Fe2+ is + 0.77 V and E° for I2/2I– = 0.536 V.The favourable redox reaction is : [2011 M]
= 0.77 – 0.536
= 0.164 V
So, reaction will taken place.
Limiting molar conductivity of NH4OH
is equal to : 
Molar conductivities at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. for CH3COOH will be : [2012 M]
= 91 + 425.9 – 126.4 = 390.5
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]
At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm-1 cm2mol-1 and at infinite dilution its molar conductance is 238 ohm-1 cm2 mol-1. The degree or ionisation of ammonium hydroxide at the same concentration and temperature is :[NEET 2013]
A button cell used in watches function s as following
If half cell potentials are :
The cell potential will be : [NEET 2013]
E°Cell = E°OP + E°RP = 0.76 + 0.314 = 1.10 V
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]
Equivalent weight of cobalt (II) = 59/2
I = 10 A
Time (t) = 109 min = 109 × 60 sec
Substituting these values we get,
Consider the half-cell reduction reaction :
The E° for the reaction and possibility of the forward reaction are, respectively [NEET Kar. 2013]
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.