Q: For the cell reaction2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(aq) ECell⊖=0.24V at 298 K. The standard Gibbs energy (△rG⊖) of the cell reaction is :[Given that Faraday constant F = 96500 C mol−1]

△G⊖=−nFEcell⊖ = – 2 × 96500 × 0.24 J mol−1 = – 46320 J mol−1 = – 46.32 kJ mol−1

Q: For a cell involving one electron Ecell0=0.59Vat 298 K, the equilibrium constant for the cell reaction is :[Given that F2.0303RT=0.059Vat T = 298 K]

Ecell=Eºcell−n0.059logQ ....(i)(At equilibrium, Q=Keq and Ecell=0 )0=Eºcell−10.059logKeq (from equation (i))logKeq=0.059Eºcell=0.0590.59=10 keq=1010=1×1010

Q: Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :Then the species undergoing disproportionation is :

=1.595−1.5=0.095 V=+ve Hence, option (D) is correct answer.

Q: The pressure of H2 required to make the potential of H2 – electrode zero in pure water at 298 k is :

From the question, we have an equation2H++2e−→H2( g)According to Nernst equationE=E0−20.0591log[H+]2PH20−20.0591log(10−7)2pH2[∴[H+]=10−7]∴ For potential of H2 electrode to zero, pH2 should be equal to [H+]2, i.e 10−14 atmlog(10−7)210−14=0

Q: The weight of silver (at wt. =108 ) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be :

According to faraday's 2nd law∴EAgwAg=EO2wO2108wAg=8(224005600)×32wAg=108 g

Q: 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?

H2⟶2H++2e− Eoxi=Eoxio−20.059logPH2[H+]2 =0−20.059log1[10−10]2 =−20.059×(−20) =0.59 V

Q: At 25∘C molar conductance of 0.1 molar aqueous solution , of ammonium hydroxide is 9.54ohm−1 cm2 mol−1 and at infinite dilution its molar conductance is 238ohm−1 cm2 mol−1 . The degree of ionisation of ammonium hydroxide at the same concentration and temperature is:

α=πm∞πm=2389.54=0.04008 or 4.008%

Question 1

On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be:

Accepted Answer

During the electrolysis of dil. sulphuric acid using Pt electrodes following reaction will take place.At cathode :		4H+(aq)+4e−⟶2H2​(g) At anode :		2H2​O(I) → O2​(g)+4H+(aq)+4e−

Question 2

The number of Faradays(F) required to produce 20 g of calcium from molten CaCl2​ (Atomic mass of Ca = 40 g mol−1) is

Accepted Answer

1 equivalent of any substance is deposited by 1 F of charge.We have, 20 g calciumNumber of equivalents = Equivalent massGiven mass​ = 2020​=1 Equivalent mass of Ca = 240​=20 So, 1 faraday of charge is required.

Question 3

For the cell reaction2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2​(aq) ECell⊖​=0.24V at 298 K. The standard Gibbs energy (△r​G⊖) of the cell reaction is :[Given that Faraday constant F = 96500 C mol−1]

Accepted Answer

△G⊖=−nFEcell⊖​ 			= – 2 × 96500 × 0.24 J mol−1 			= – 46320 J mol−1 			= – 46.32 kJ mol−1

Question 4

For a cell involving one electron Ecell0​=0.59Vat 298 K, the equilibrium constant for the cell reaction is :[Given that F2.0303RT​=0.059Vat T = 298 K]

Accepted Answer

Ecell​=Eºcell​−n0.059​logQ 		....(i)(At equilibrium, Q=Keq​ and Ecell​=0 )0=Eºcell​−10.059​logKeq​ (from equation (i))logKeq​=0.059Eºcell​​=0.0590.59​=10 keq​=1010=1×1010

Question 5

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :Then the species undergoing disproportionation is :

Accepted Answer

​=1.595−1.5=0.095 V=+ve​ Hence, option (D) is correct answer.

Question 6

The pressure of H2​ required to make the potential of H2​  – electrode zero in pure water at 298 k is :

Accepted Answer

From the question, we have an equation2H++2e−→H2​( g)According to Nernst equationE=E0−20.0591​log[H+]2PH2​​​0−20.0591​log(10−7)2pH2​​​[∴[H+]=10−7]∴ For potential of H2​ electrode to zero, pH2​ should be equal to [H+]2, i.e 10−14 atmlog(10−7)210−14​=0

Question 7

The weight of silver (at wt. =108 ) displaced by a quantity of electricity which displaces 5600 mL of O2​ at STP will be :

Accepted Answer

According to faraday's 2nd   law∴​EAg​wAg​​=EO2​​wO2​​​108wAg​​=8(224005600​)×32​wAg​=108 g​

Question 8

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?

Accepted Answer

H2​⟶2H++2e− Eoxi​=Eoxio​−20.059​logPH2​​[H+]2​ =0−20.059​log1[10−10]2​ =−20.059​×(−20) =0.59 V

Question 9

At 25∘C  molar conductance of 0.1  molar aqueous solution , of ammonium hydroxide is 9.54ohm−1 cm2 mol−1  and at infinite dilution its molar conductance is 238ohm−1 cm2 mol−1 . The degree of ionisation of ammonium hydroxide at the same concentration and temperature is:

Accepted Answer

α=πm∞​πm​​=2389.54​=0.04008 or 4.008%

Question 10

A button cell used in watches functions as following​Zn(s)+Ag2​O(s)+H2​O(l)⇌2Ag(s)+Zn2+(aq)+2OH−(aq)​ If half cell potentials are​Zn2+(aq)+2e−→Zn(s);E∘=−0.76 VAg2​O(s)+H2​O(l)+2e−→2Ag(s)+2OH−(aq)Eo=0.34 V​ The cell potential will be:

Accepted Answer

Cell rep Zn/Zn+2∥Ag+∣Ag EoCell=0.76+(0.34)=1.1 V

Question 11

The correct order of the stoichiometries of AgCl formed when AgNO3​  in excess is treated with the complexes : CoCl3​⋅6NH3​,CoCl3​⋅5NH3​,CoCl3​−4NH3​  respectively is:

Accepted Answer

Complexes are respectively [Co(NH3​)6​]Cl3​,[Co(NH3​)5​Cl]Cl2​  and [Co(NH3​)4​Cl2​]Cl

Question 12

Arrange the following metals in the order in which they displace each other from the solution of their salts.Al,Cu,Fe,Mg and Zn

Accepted Answer

The order in which the metals displace each other from the solution of their salts is Mg,Al,Zn,Fe,Cu. Thus, more electropositive Mg can displace less electropositive Cu.

Question 13

Given the standard electrode potentials, K+/K=−2.93 V,Ag+/Ag=0.80 V, Hg2+/Hg=0.79 V, Mg2+/Mg=−2.37V,Cr3+/Cr=−0.74 VArrange these metals in their increasing order of reducing power.

Accepted Answer

The increasing order of reducing power is as follows:Ag+∣Ag<Hg2+∣Hg<Cr3+∣Cr< Mg2+∣ Mg<K+∣KMetal with most positive standard electrode potential has the least reducing power.Metal with most negative standard electrode potential has maximum reducing power.

Question 14

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place. Further show:(i) Which of the electrode is negatively charged?(ii) The carriers of the current in the cell.(iii) Individual reaction at each electrode.

Accepted Answer

The galvanic cell in which the given reaction takes place is depicted as: Zn(s)∣Zn2+(aq)∣∣Ag+(aq)∣Ag(s)(i) The negatively charged electrode is the zinc electrode. It acts as an anode.(ii) In the external circuit, the current will flow from silver to zinc. (iii) Oxidation at anode: Zn(s)→Zn2+(aq)+2e−Reduction at cathode: Ag+(aq)+e−→Ag(s)

Question 15

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:(i) 2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd(ii) Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Calculate the Δr​G⊖ and equilibrium constant of the reactions.

Accepted Answer

(i) Ecello​=Ecathode0​−Eanode0​=−0.40−(−0.74)=0.34 VΔG0=−nFEcello​=−6×96500×0.34=196860J/mol=196.86kJ/mollog Kc​=0.059nEcello​​=0.0596×0.34​=34.576log Kc​=0.059nEcello​​=0.0596×0.34​=34.576(ii) Ecello​=Ecathode0​−Eanode0​=0.80−(−0.77)=0.03 VKc​=3.76×1034log Kc​=0.059nEcello​​=0.0591×0.03​=0.508Ecello​=Ecathode0​−Eanode0​=0.80−(−0.77)=0.03

Question 16

Write the Nernst equation and emf of the following cells at 298 K:(i) Mg(s)∣Mg+2(0.001 M)∥Cu+2(0.0001 M)∣Cu(s).(ii) Fe(s)∣Fe+2(0.001 M)∥H+(1M)∣H2​(g)(1bar)∣Pt(s)(iii) Sn(s)∣Sn2+(0.050 M)∥H+(0.020 M)∣H2​(g)(1bar)∣Pt(s)(iv) Pt(s)∣Br2​(l)∣Br−(0.010 M)∥H+(0.030 M)∣H2​(g)(1bar)∣Pt(s).

Accepted Answer

(i) Eo=0.34−(−2.37)=2.71 VE=E0−n0.059​log[Cu2+][Mg2+]​E=2.71−20.059​log(0.0001)0.001​E=2.6805 V(ii) Eo=0−(−0.44)=0.44 VE=E0−n0.059​log[H+]2[Fe2+]​E=0.44−20.059​log(1)20.001​E=0.5285 V(iii) Eo=0−(−0.14)=0.14 VE=E0−n0.059​log[H+]2[Sn2+]​E=0.14−20.059​log(0.020)20.050​E=0.08 V(iv) Eo=0−1.08=−1.08 VE=E0−20.059​log[Br−]2[H+]21​E=−1.08−20.059​log(0.01)2(0.030)21​E=−1.288 V

Question 17

In the button cells widely used in watches and other devices the following reaction takes place:Zn(s)+Ag2​O(s)+H2​O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)Determine Δr​Gθ and Eθ for the reaction.

Accepted Answer

Ecell0​=Ecathode0​−Eanode0​=EAgo​−EZno​=0.344−(−0.76)=1.104 VΔG0=−nFEcell0​=−2×96500×1.104=−213072J=−213kJ

Question 18

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Accepted Answer

Conductivity (K):It is the conductance of unit cube of material. S.I unit is S/m. Common unit is S/cm.The conductivity of an electrolytic solution always decreases with decrease in concentration that is on dilution. This is because with dilution, the degree of dissociation increases and the total number of current-carrying ions increases but the number of ions per unit volume decreases.Molar conductivity Λ:It is the ratio of the electrolytic conductivity k to the molar concentration C of the dissolved electrolyte. Λ=Ck​It is also defined as the conductance of a volume of solution containing 1 mole of dissolved electrolyte when placed between parallel electrodes 1 cm apart and large enough to contain between them all the solution.The S.I unit of molar conductivity is  Sm2/molThe common unit of molar conductivity is  Scm2/molThe molar conductivity of strong and weak electrolytes increases with dilution (i.e, decrease in the concentration). This is because with dilution, the degree of dissociation increases and the total number of current-carrying ions increases.

Question 19

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1. Calculate its molar conductivity.

Accepted Answer

Molar conductivity Λ and specific conductivity κ are related by the expression as follows:Λ=C1000κ​where C is the molar concentration.Λ=0.201000×0.0248​=124.0S cm2/mol

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Solution for Class 12, Chemistry, Electrochemistry

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