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Conductivity (unit Siemen’s S) is directly proportional toarea of the vessel and the concentration of the solution in itand is inversely proportional to the length of the vesselthen the unit of the constant of proportionality is [2002]
given :
EMF of a cell in terms of reduction potential of its left andright electrodes is [2002]
E_{cell} = Reduction potential of cathode (right)
– Reduction potential of anode (left)
= E_{right} – E_{left.}
What will be the emf for the given cell [2002]
Pt  H_{2} (P_{1})  H^{+} (aq)   H_{2} (P_{2})  Pt
Oxidation half call:
The net cell reaction
Which of the following reaction is possible at anode?
O.S. of Cr changes from +3 to +6 by loss of electrons. At anode oxidation takes place.
When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are
[2002]
Pure metal always deposits at cathode
Which of the following is a redox reaction? [2002]
For a cell reaction involving a twoelectron change, thestandard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25ºC will be [2003
The equilibrium constant is related to the standard emf of cell by the expression
Standard reduction electrode potentials of three metals A, B& C are respectively + 0.5 V, – 3.0 V & –1.2 V. The reducingpowers of these metals are [2003]
NOTE : The higher the negative value of reduction potential, the more is the reducing power.
Hence B > C > A.
When during electrolysis of a solution of AgNO_{3} 9650coulombs of charge pass through the electroplating bath,the mass of silver deposited on the cathode will be
When 96500 coulomb of electricity is passed through the electroplating bath the amount of Ag deposited = 108g
∴ when 9650 coulomb of electricity is passed deposited Ag.
For the redox reaction :
Zn(s) +Cu^{2+} (0.1M)→ Zn^{2+} (1M) + Cu(s) taking place in a cell, is 1.10 volt. E_{cell} for the cell will be [2003]
Several blocks of magnesium are fixed to the bottom of a ship to [2003]
Magnesium, on account of its lightness, great affinity for oxygen and toughness is used in ship. Being a lighter element, magnesium makes the ship lighter when it is fixed to the bottom of the ship.
In a hydrogenoxygen fuel cell, combustion of hydrogenoccurs to [2004]
In H_{2} O_{2} fuel cell, the combustion of H_{2} occurs to create potential difference between the two electrodes
Consider the following Eº values
Under standard conditions the potential for the reaction
[2004]
Fe^{3+} + e^{} → Fe^{2+}ΔG° = 1x Fx0.77
Sn^{2+} + 2e^{} → Sn(s)ΔG° = 2 x F(0.14)
for Sn(s) + 2Fe^{3+} (aq) → 2Fe^{2+} (aq) +Sn^{2+} (aq)
∴ Standard potential for the given reaction
∴
The standard e.m.f. of a cell involving one electron changeis found to be 0.591 V at 25ºC. The equilibrium constant ofthe reaction is (F = 96,500 C mol^{–1}; R = 8.314 JK^{–1} mol^{–1}) [2004]
The limiting molar conductivities Lº for NaCl, KBr and KClare 126, 152 and 150 S cm^{2} mol^{–1} respectively. The Lº forNaBr is [2004]
operating (i) + (ii)  (iii)
= 126 +152 150 = 128 S cm^{2} mol^{1}
In a cell that utilises the reaction Zn_{(s)} + 2H^{+} (aq) → Zn^{2+}+ H_{2(g)} addition of H_{2}SO_{4 }to cathode compartment, will [2004]
The values for Cr, Mn, Fe and Co are – 0.41, + 1.57, + 0.77 and + 1.97V respectively. For which one of these
metals the change in oxidation state from +2 to +3 is easiest? [2004]
The given values show that Cr has maximum oxidation potental, therefore its oxidation will be easiest. (Change the sign to get the oxidation values)
For a spontaneous reaction the DG, equilibrium constant (K) and will be respectively [2005]
NOTE : For spontaneous reaction DG should be negative. Equilibrium constant should be more than one
(ΔG = – 2.303 RT log K_{c}, If K_{c} = 1 then ΔG = 0; If K_{c} < 1
The highest electrical conductivity of the following aqueoussolutions is of [2005]
Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid > fluoro acetic acid > chloro acitic acid > acetic acid.
Aluminium oxide may be electrolysed at 1000°C to furnishaluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500Coulombs). The cathode reaction is– Al^{3+} + 3e^{} → Al° To prepare 5.12 kg of aluminium metal by this method werequire
1 mole of e– = 1F = 96500 C
27g of Al is deposited by 3 × 96500 C
5120 g of Al will be deposited by
Calculate using appropriate molar conductances of the electrolytes listed above at infinite dilution in H_{2}O
at 25°C
Which of the following chemical reactions depict the oxidizing beahviour of H_{2}SO_{4}?
reaction oxidation number of S is decreasing from + 6 to +4 hence undergoing reduction and for HI oxidation
Number of I is increasing from –1 to 0 hence
underegoing oxidation therefore H_{2}SO_{4} is acting as oxidising agent.
The molar conductivities and at infinite dilution in water at 25ºC are 91.0 and 426.2 S cm2/mol
respectively. To calculate , the additional value required is
is given by the following equation
Hence is required.
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 W. The conductivity of this solution is 1.29 S m^{–1}. Resistance of the same cell when filled with 0.2 M of the same solution is 520 W. The molar conductivity of 0.2 M solution of electrolyte will be [2006]
Given, R = 520Ω, C = 0.2 M, μ (molar conductivity) = ?
now cell constant is known.)
Hence,
The equivalent conductances of two strong electrolytes at infinite dilution in H_{2}O (where ions move freely through a
solution) at 25°C are given below : [2007]
What additional information/ quantity one needs to calculate of an aqueous solution of acetic acid?
NOTE : According to Kohlrausch’s law, molar
conductivity of weak electrolyte acetic acid (CH_{3}COOH) can be calculated as follows:
∴ Value of should also be known for
calculating value of
The cell, Zn  Zn^{2+} (1 M)  Cu^{2+} (1 M)  Cu was allowed to be completely discharged at 298 K. The relative
concentration of is [2007]
E_{cell} = 0; when cell is completely discharged.
Given The potential for the cell CrCr^{3+} (0.1M) Fe^{2+} (0.01 M) Fe is [2008]
From the given representation of the cell, E_{cell} can be found as follows.
Hence option (d) is correct answer.
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
At 298 K standard Gibb’s energies of formation for CH_{3}OH(l), H_{2}O(l) and and CO_{2} (g) are –166.2, –237.2 and –394.4 kJ
mol^{–1} respectively. If standard enthalpy of combustion of methonal is – 726 kJ mol^{–1}, efficiency of the fuel cell will be:
[2009]
= – 394.4 + 2 (–237.2) – (–166.2) – 0
= – 394.4 – 474.4 + 166.2 = – 702.6 k J
% efficiency =
Given:
The value of standard electrode potential for the change, Fe^{3+} (aq) + e^{–} →Fe^{2+} (aq) will be:
[2009]
Given
we have to calculate
Fe^{3+} + e^{–} → Fe^{2+} … (iii)
As we know that ΔG = –nFE
Thus for reaction (iii)
ΔG = ΔG_{1}  ΔG ; – nFE° = – nFE_{1} – (–nFE_{2})
–nFE° = nFE_{2} – nFE_{1}
–1FE° = 2× 0.439F – 3 × 0.036 F
–1 FE° = 0.770 F ∴ E° = – 0.770V
The Gibbs energy for the decomposition of Al_{2}O_{3} at 500°C is as follows :
The potential difference needed for electrolytic reductionof Al_{2}O_{3} at 500°C is at least [2010]
The correct order of values with negative sign for the four successive elements Cr, Mn, Fe and Co is [2010]
The value of for given metal ions are
The correct order of values without
considering negative sign would be
Mn^{2+} > Cr^{2+} > Fe^{2+} > Co^{2+}.
The reduction potential of hydrogen halfcell will be negative if :
[2011]
Now if = 2 atm and [H^{+}] = 1M
The standard reduction potentials for Zn^{2+}/Zn, Ni^{2+}/Ni and Fe^{2+}/Fe are –0.76,–0.23 and –0.44 V respectively.
The reaction X+Y^{2+} →X^{2+}+Y will be spontaneous when :
For a spontaneous reaction DG must be –ve
Since ΔG = – nFE°
Hence for ΔG to be ve ΔE° has to be positive. Which is possible when X = Zn, Y = Ni
Zn + Ni^{++} → Zn^{++} + Ni
Given :
Based on the data given above, strongest oxidising agentwill be : [JEE M 2013]
higher the value of standard reduction potential
stronger will be the oxidising agent, hence is the strongest oxidising agent.
Resistance of 0.2 M solution of an electrolyte is 50 Ω.The specific conductance of the solution is 1.4 S m^{–1}.The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m^{2} mol^{–1} is:
[JEE M 2014]
Given for 0.2 M solution
For 0.5 M solution
R = 280 Ω; k = ?
= 5 S cm^{2} mol^{–1} = 5 × 10^{–4} S m^{2} mol^{–1}
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