set-3 “delhi” class xii-science : chemistry, board paper 2014,studymoz.com/files/exam...
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Class XII-Science : Chemistry, Board Paper 2014,Set-3 “Delhi”
Ge ne ral Instructio ns:Ge ne ral Instructio ns:
(i) All questions are compulsory
(ii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each.
(iii) Question numbers 9 to 18 are short-answer questions and carry 2 marks each.
(iv) Question numbers 19 to 27 are also short-answer questions and carry 3 marks.
(v) Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
(vi) Use Log Tables, if necessary. Use of calculators is not allowed.
Question 1
Question 1
Questions
Q 1Q 1
Give one example each of lyophobic sol and lyophilic sol.
Solution:
(i) Lyophobic sols: Gum, gelatin, starch, proteins, rubber etc. (any one)(ii) Lyophilic sols: Solutions of metals such as silver and gold, iron (III) hydroxide solution (Fe(OH)3), arsenic sulphide (As2S3)solution etc. (any one)Q 2Q 2
Write the IUPAC name of the compound.
Solution:
The IUPAC name of the given compound is 4-Hydroxypentan-2-one.
Q 3Q 3
What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Solution:
Dipole-dipole intermolecular attractive interaction exists between methanol and acetone, as both are polar molecules.
Q 4Q 4
Which of the following is a more stable complex and why ?
CoNH363+ and Coen33+
Solution:
Chelating ligands form more stable complexes compared to non-chelating ligands. Since ethylene diammine is a
bidentate ligand and forms stable chelate, [Co(en)3]3+ will be a more stable complex than [Co(NH3)6]
3+.
Q 5Q 5
Arrange the following in increasing order of basic strength :
C6H5NH2, C6H5NHCH3, C6H5N(CH3)2
Solution:
Increasing order of basic strength in gaseous state is as follows:
C6H5NH2 < C6H5NHCH3 < C6H5N(CH3)2
Q 6Q 6
Name the products of hydrolysis of sucrose.
Solution:
Products of hydrolysis of sucrose are D-(+) glucose and D-(-)-fructose.
Q 7Q 7
Which of the following isomers is more volatile:
o-nitrophenol or p-nitrophenol?
Solution:
o-nitrophenol forms intra-molecular hydrogen bond. On the other hand, p-nitrophenol is involved in stronger inter-molecular hydrogen bonding and hence, it has higher
boiling point than o-nitrophenol. So, o-nitrophenol is more volatile than p-nitrophenol.
Q 8Q 8
Which reducing agent is employed to get copper from the leached low-grade copper ore?
Solution:
Copper can be obtained from low-grade ore through the process of leaching using acid or bacteria (leaching is a process in
which ore is treated with suitable reagent that dissolves ore but not the impurities).
The solution containing copper can be reduced with the help of reducing agents such as scrap iron or H2 to get copper metal.
Cu2+(aq) + Fe ® Cu(s) + Fe2+(aq)
Cu2+ (aq) + H2 (g) ® Cu(s) + 2 H+(aq)
Q 9Q 9
State Raoult's law for the solution containing volatile components. What is the similarity between Raoult's law and Henry's
law?
Solution:
Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly
proportional to its mole fraction.
Thus, if there is a solution of two liquid components (1 and 2), then for component 1:
p1 ∝ x1
p1=p1o x1
For component 2:
p2=p2o x2
Here:
p1, p2 = Partial vapour pressure of two volatile components (1 and 2) of the solution
p1o, p2o
= Vapour pressure of pure components (1 and 2)
x1 , x2 = Mole fractions of the components (1 and 2)
Similarity between Raoult's law and Henry's law:
Both laws state that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in the
solution.
Q 10Q 10
Explain the following terms :
(i) Rate constant (k)
(ii) Half life period of a reaction (t1/2)
Solution:
(i) R ate Co nstantR ate Co nstant
Rate constant is defined as the rate of a reaction when concentration of the reactants is unity. It is defined by k.
Unit of k is mol1−nLn−1s−1, where n is the order of the reaction.
(ii) Half life pe rio d o f a re actio n (Half life pe rio d o f a re actio n ( tt 11 /2/2 ))
The half life of a reaction is the time period in which the concentration of a reactant is reduced to one half of its initial
concentration.
Q 11Q 11
Write the principles of the following methods :
(i) Froth floatation method
(ii) Electrolytic refining
Solution:
( i) Fro th Flo atatio n Me tho d:( i) Fro th Flo atatio n Me tho d: This method is based on the principle that the ore particles are wetted by oil, whereas gangue
particles are wetted by water. This method is used for the concentration of sulphide ores.
( ii) Ele ctro lytic R e fining:( ii) Ele ctro lytic R e fining: In this method, impure metal is refined using electricity. The impure metal is made the anode and a
strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. Impurities get
collected below the anode and are known as anode mud.
At anode, the metal looses electrons and forms a positively charged metal ion and comes in the solution.
M ® Mn+ + ne-
At cathode, the metal ion gains electrons and gets deposited on the electrode.
Mn+ + ne- ® M
Q 12Q 12
An element with density 11.2 g cm–3 forms a f.c.c. lattice with edge length of 4 × 10–8 cm.
Calculate the atomic mass of the element.
(Given : NA = 6.022 × 1023 mol–1)
Solution:
Given,
Density, d = 11.2 g cm-3
Edge length, a = 4 10-8 cm
Avogadro number, NA = 6.022 X 1023
Number of atoms present per unit cell, Z (fcc) = 4
We know for a crystal system,
d=z×Ma3×NAÞM=d×a3×NAZÞM =11.2×64×10-24 ×6.022×10234=107.91 g
�
Thus, atomic mass of the element is 107.91 g.
Q 13Q 13
Examine the given defective crystal:
Answer the following questions :
(i) What type of stoichiometric defect is shown by the crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect?
Solution:
(i) Schottky defect is shown by the mentioned crystal, as equal number of cations and anions are missing in the crystal lattice.
(ii) This defect leads to decrease in density, as equal number of the cations and anions are missing from the crystal lattice.
(iii) This kind of defect is shown by those ionic substance in which the cations and anions are of almost similar sizes.
Examples: NaCl, KCl and CsCl.
Q 14Q 14
Calculate the mass of a compound (molar mass = 256 g mol−1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf = 5.12 K kg mol−1).
Solution:
Given:ΔTf =0.48 KKf=5.12 K kg mol-1w1= 75 gw2=?M2=256 g mol-1UsingΔTf =Kf×w2×1000M2×w1Þw2=ΔTf×M2×w1Kf×1000
Q 15Q 15
(i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction?
(ii) Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation
Solution:
(i)
Among the given pair of compounds, alkyl halide (b) has a chiral centre.
The alkyl halide (a) does not contain a chiral centre and and it also gives faster SN2 reaction as SN2 is more favourable in primary alkyl halides.
(ii)
(a) Inversion of configuration takes place in SN2 reaction.
(b) Racemisation takes place in SN1 reaction.
Q 16Q 16
Draw the structure of major monohalo product in each of the following reactions :
(i)
(ii)
Solution:
(i)
(ii) Addition in presence of peroxide yields product according to anti-Markovnikov rule of addition.
Q 17Q 17
Complete the following chemical equations :
(i) Ca3P2 + H2O ®
(ii) Cu + H2SO4(conc.) ®
OR
Arrange the following in the order of the property indicated against each set :
(i) HF, HCl, HBr, HI − increasing bond-dissociation enthalpy.
(ii) H2O, H2S, H2Se, H2Te − increasing acidic character.
Solution:
The balanced reactions are given below:
i) Ca3P2 + 6H2O ®3Ca(OH)2 + 2PH3
ii) Cu + 2H2SO4(conc.) ® CuSO4 + SO2 + 2H2O
OR
i)
The arrangement of the given hydrogen halides in increasing order of bond-dissociation enthalpy is given below:
H−I < H−Br < H−Cl < H−F
ii)
The increasing order of acidic character of the given hydrides of Group 16 elements is given below:
H2O < H2S < H2Se < H2Te
Q 18Q 18
Write the IUPAC name of the complex [Cr(NH3)4Cl2]+. What type of isomerism does it exhibit?
Solution:
The IUPAC name of the complex [Cr(NH3)4Cl2]+ is Tetraamminedichlorochromium(III) ion.
This complex exhibits geometrical isomerism. [Cr(NH3)4Cl2]+ is a [MA4B2] type of complex, in which the two chloride ligands may be orientedcis and trans to each
other.
Q 19Q 19
(a) Write the mechanism of the following reaction :
CH3CH2OH ® HBr CH3CH2Br+H2O
(b) Write the equation involved in Reimer-Tiemann reaction.
Solution:
a.The reaction proceeds through substitution nucleophilic bimolecular (SN2) mechanism, as shown:
Inversion of configuration takes place during the reaction.
b.b.
R e ime r-Tie mann re actio n:R e ime r-Tie mann re actio n:
Q 20Q 20
(a) Draw the structures of the following compounds :
(i) XeF4
(ii) N2O5
(b) Write the structural difference between white phosphorous and red phosphorous.
Solution:
(a)
(i) The structure of XeF4 is given below:
(ii) The structure of N2O5 is given below:
(b) Structural difference between white P and red P:
White P Red P
It consists offour P atoms, linkedwith one another to giverise to a tetrahedralshape.
It has a polymericstructure, consisting ofchains of P4 tetrahedral
units that are linkedtogether.
Structure: Structure:
Q 21Q 21
Give the structures of A, B and C in the following reactions :
(i)
(ii)
OR
How will you convert the following?
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N−phenylethanamide
(Write the chemical equations involved.)
Solution:
(i)
(ii)
OR(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N−phenylethanamide
Q 22Q 22
Account for the following :
(i) Sulphur in vapour form exhibits paramagnetic behaviour.
(ii) SnCl4 is more covalent than SnCl2.
(iii) H3PO2 is a stronger reducing agent than H3PO3.
Solution:
(i)
Sulphur partly exists as S2 molecule in vapour state, which has two unpaired electrons in the antibonding
p*
molecular orbitals and hence, it exhibits paramagnetism.
(ii)
According to the Fajan's rule, the central metal with more positive oxidation number, shows more polarising power. This in turn
shows more covalent character. In SnCl4, Sn exists in +4 oxidation state, while in SnCl2, Sn exists in +2 oxidation state. So, SnCl4
is more covalent than SnCl2.
(iii)
Greater the number of element-hydrogen (E−H) bonds present in a compound, greater is the reducing power of the
compound. H3PO2 is a stronger reducing agent as it contains 2 P−H bonds, while H3PO3 has only one P−H bond.
Q 23Q 23
(i) What are disinfectants? Give an example.
(ii) Give two examples of macromolecules that are chosen as drug targets.
(iii) What are anionic detergents? Give an example.
Solution:
(i) Disinfectants: A disinfectant is used primarily to sanitise objects. It acts as an antimicrobial that kills the microbes that cause
infection. Disinfectants usually come as liquids or sprays.
Example: Chlorine in the concentration of 0.2 to 0.4 ppm, in an aqueous solution, acts as a disinfectant.
( ii)( ii) Enzymes and receptors are the macromolecules that are chosen as drug targets.
( iii) Anio nic de te rge nts:( iii) Anio nic de te rge nts: The detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons.
Anionic detergents are of two types:1. Sodium alkyl sulphates: Sodium lauryl sulphate (C11H23CH2OSO3
−Na+) and sodium stearyl sulphate (C17H35CH2OSO3−Na+).
2. Sodium alkylbenzenesulphonates: Sodium 4-(1-dodecy) benzenesulphonate (SDS)Q 24Q 24
(i) Deficiency of which vitamin causes scurvy?
(ii) What type of linkage is responsible for the formation of proteins?
(iii) Write the product formed when glucose is treated with HI.
Solution:
( i)( i)
Deficiency of vitamin C causes scurvy. Bleeding gums is a symptom of scurvy.
( ii)( ii)
Peptide linkage is responsible for the formation of proteins. Protein is formed by the combination of the amino acids that
contain amino (-NH2) and carboxyl(-COOH) as functional groups.
( iii)( iii)
When glucose is treated with HI, n-Hexane is formed as the final product.
Reaction involved:
OHC-(CHOH)4-CH2OH + HI
®CH3 -CH2-CH2-CH2-CH2-CH3
(Glucose) (n-Hexane)
Q 25Q 25
(a) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(b) Write an important characteristic of lyophilic sols.
(c) Based on the type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid.
Solution:
(a)Freundlich adsorption isotherm for adsorption of gases on solids:xm=Kp1nIt can also be written aslogxm=log K + 1nlog pwhere x is the mass of the adsorbate, m is the mass of the absorbent and p is the pressure of the gas and n is a constant which isgreater than 1.(b)
Lyophilic sols are sols that are solvent-attracting. An important characteristic of these sols is that if the dispersion medium isseparated from the dispersion phase by any method, the sol can be reconstituted by simply remixing the two again. That is whythese sols are also known as reversible sols.(c)
Example of associated colloid: Soap solution
Example of multimolecular colloid: Gold sol
Q 26Q 26
The following data were obtained during the first-order thermal decomposition of SO2Cl2 at a constant volume:
SO2Cl2(g) ® SO2(g) + Cl2(g)
Experiment Time/s−1 Totalpressure/atm
12
0100
0.40.7
Calculate the rate constant.
(Given : log 4 = 0.6021, log 2 = 0.3010)
Solution:
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation:
After time t, total pressure,
p=P0+pp=Pt-P0Therefore, = 2 P0 − Pt
For a first-order reaction,
When t = 100 s,k=2.303100log0.42×0.4-0.7= 1.387×10-2 s-1
Q 27Q 27
After the ban on plastic bags, students of a school decided to make people aware of the harmful effects of plastic bags on the environment and Yamuna River. To make
the awareness more impactful, they organised a rally by partnering with other schools and distributed paper bags to vegetable vendors, shopkeepers and departmental
stores. All the students pledged not to use polythene bags in the future to save the Yamuna River.
After reading the above passage, answer the following questions:
(i) What values are shown by the students?
(ii) What are bio-degradable polymers? Give one example.
(iii) Is polythene a condensation or an addition polymer?
Solution:
(i) From the given passage, we can conclude that the students show awareness about the environment.
(ii) A polymer that can be decomposed by microorganisms within a definite period of time, so that the polymer or its degraded
product does not cause any harm to the environment, is called a bio-degradable polymer.For example, poly-b-hydroxybutyrate-CO-b- hydroxy valerate (PHBV) is a bio-degradable aliphatic polyester.
(iii) Polythene is an addition polymer that is formed by addition of ethene molecules.
Q 28Q 28
(a) How do you prepare:
(i) K2MnO4 from MnO2?
(ii) Na2Cr2O7 from Na2CrO4?
(b) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) The enthalpy of atomisation is lowest for Zn in 3d series of the transition elements.
(iii) Actinoid elements show wide range of oxidation states.
OR
(i) Name the elements of 3d transition series that show maximum number of oxidation states. Why does this happen?
(ii) Which transition metal of 3d series has positive E° (M2+/M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidising agent and why?
(iv) Name a member of the lanthanoid series that is well-known to exhibit +2 oxidation state.
(v) Complete the following equation:
MnO4-+8H++5e- ®
Solution:
a)
i) K2MnO4 can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an
oxidising agent, such as KNO3 or KClO4, to give K2MnO4.
2 MnO2 + 4 KOH + O2
®Δ
2 K2MnO4 +2 H2O
green
�ii)
Na2Cr2O7 can be prepared from Na2CrO4 in the following way:
For the preparation of sodium dichromate, the yellow solution of sodium chromate is acidified with sulphuric acid to give a
solution from which orange sodium dichromate, Na2Cr2O7.2H2O can be crystallised.
Balanced equation for above reactions is as follows:
2 Na2CrO4 + 2 H+
® Na2Cr2O7 + 2 Na+ + H2O
Yellow Orange
�b)
i) Electronic configuration of Mn2+ is [Ar]18 3d5 .
Electronic configuration of Fe2+ is [Ar]18 3d6 .
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in +2 state has a stable d5 configuration.
Therefore, Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its
configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ gets oxidised to Fe3+ easily.
ii) The extent of metallic bonding an element undergoes, decides the enthalpy of atomisation. The more extensive the metallic
bonding of an element, the more will be its enthalpy of atomisation. In all transition metals (except Zn, electronic configuration:
3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these
unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of
atomisation.
iii) Actinides exhibit larger oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells . The energies
are calculated on the basis of (n+l) rule. The (n+l) values of the three orbitals are:
5 f = 5 + 3 = 8
6 d = 6 + 2 = 8
7 s = 7 + 0 = 7
Since, all the values are almost same, therefore all orbitals can involved in bonding resulting in larger oxidation number for
actinoids.
OR
1)
In 3d-series of transition metals, manganese has an atomic number of 25 that gives the electronic configuration as [Ar] 3d54s2
,where we see that the maximum number of unpaired electrons is found in manganese atom; so, it can show a maximum
oxidation state upto +7.
2)
Copper is the transition metal of 3d series that exhibits positive E0(M2+/M). The value of E0(M2+/M) for copper is (+0.34). This
happens because the E0(M2+/M) value of a metal depends on the energy changes involved in the following:
1. Sublimation energy: The energy required for converting one mole of an atom from the solid state to the gaseous state.
M(s)
®M(g) Δs H (Sublimation energy)
2. Ionisation energy: The energy required to take out electrons from one mole of atom in the isolated gaseous state.
M(g)
®
M2+(g) ΔiH (Ionisation energy)
3. Hydration energy: The energy released when one mole of ions are hydrated.
M2+(g)
®
M2+(aq) ΔhydH (Hydration energy)
Since, copper has a high energy of atomisation and low hydration energy, the E0(M2+/M) value for copper is positive.
3) Out of Cr3+ and Mn3+, Mn3+ is a stronger oxidising agent because it has 4 electrons in its valence shell and when it gains one
electron to form Mn2+, it results in the half-filled (d5) configuration that has extra stability.
4) Europium (Eu) is well-known to exhibit +2 oxidation state due to its half filled f orbital in +2 oxidation state.
5) MnO4- + 8H+ + 5e-
®
Mn2+ + 4 H2O
Q 29Q 29
(a) Write the products of the following reactions:
(i)
(ii)
(iii)
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
O RO R
(a) Account for the following:
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
(ii) Carboxylic acid is a stronger acid than phenol.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Wolff-Kishner reduction
(ii) Aldol condensation
(iii) Cannizzaro reaction
Solution:
(a)
(i)
(ii)
(iii)
(b)
(i) Benzaldehyde and Benzoic acidTest-1 Through sodium bicarbonateBenzaldehyde does not react with sodium bicarbonate. However, benzoic acid will produce brisk effervescence on reaction withsodium bicarbonate as shown in the given reaction:C6H5COOH + NaHCO3
®C6H5COONa + H2O + CO2
Test-2 Through Tollen's reagentBenzaldehyde reacts with ammoniacal solution of silver nitrate to form a silver mirror.
C6H5CHO + 2[Ag(NH3)2]+ + 3OH−
®
C6H5COO− + 2Ag + 2H2O + 4NH3
However, no such reaction is given by benzoic acid. (ii)Test-1 Iodoform TestPropanone gives positive iodoform test, as it contains CH3CO group, whereas propanal does not give iodoform test. The reactionis as follows:
CH3CH2CHO + 3I2 + 4NaOH ® No reactionHeatPropanal
OR(a)
(i)
CH3COCH3 is sterically hindered than CH3CHO due to the presence of alkyl group on both sides of the carbonyl carbon, making
them less reactive towards nucleophilic attack because both methyl groups have electron releasing tendency due to -I effect.
These alkyl groups make ketone less reactive by donating electron to carbonyl group. Therefore, acetaldehyde is more reactive
towards reaction with HCN.
(ii)
Carboxylic acids are acidic due to resonance stabilisation of carboxylate anion and in phenols, acidic character is present due
to resonance stabilisation of phenoxide anion. Carboxylic acids are more acidic than phenols because the negative charge in
carboxylate anion is more spread out compared to the phenoxide ion, as there are two electronegative O-atoms in
carboxylate anion compared to one in phenoxide ion. In the resonance structures of carboxylate anion, the negative charge
is present on the O-atoms, while in resonance of phenoxide ion, negative charge is also present on electropositive carbon
atom, which leads to less stability of phenoxide ion than carboxylate anion.
(b)
Q 30Q 30
(a) Define the following terms :
(i) Limiting molar conductivity
(ii) Fuel cell
(b) Resistance of a conductivity cell filled with 0.1 mol L−1 KCl solution is 100 W. If the resistance of the same cell when filled with
0.02 mol L−1 KCl solution is 520 W, calculate the conductivity and molar conductivity of 0.02 mol L−1 KCl solution. The
conductivity of 0.1 mol L−1 KCl solution is 1.29 × 10−2 W−1 cm−1.
O RO R
(a) State Faraday's First Law of Electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+
to Cu.
(b) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given Eoce ll = +2.71 V, 1 F = 96500 C mol–1]
Solution:
(a)
(i) When concentration of an electrolyte approaches zero, then its molar conductivity is known as limiting molar conductivity.
(ii)
Fuel cells are the galvanic cells in which the energy of combustion of the fuels like hydrogen, methanol. etc is directly
converted into electrical energy.
(b)
Given that:
Concentration of the KCl solution = 0.1 mol L−1
Resistance of cell filled with 0.1 mol L−1 KCl solution = 100 ohm
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Cell constant = G* = conductivity
×
resistance
1.29×10−2 ohm−1cm−1 × 100 ohm = 1.29 cm−1 = 129 m−1
Cell constant for a particular conductivity cell is a consant.
Conductivity of 0.02 mol L−1 KCl solution =
Cell constantResistance
=
G*R=129 m-1520 ohm=0.248 Sm-1
Concentration = 0.02 mol L−1
= 1000
×
0.02 mol m−3 = 20 mol m−3
Now,
Molar conductivity =
Lm=kc=248×10-3 S m-120 mol m-3=124×10-4 S m2mol-1
Therefore, the molar conductivity of 0.02 mol L−1 KCl solution is
124×10-4 S m2 mol-1
.
O RO R
(a)
Faraday's first law o f e le ctro lysis Faraday's first law o f e le ctro lysis states that "the amount of chemical reaction which occurs at any electrode during
electrolysis by a current is proportional to the quantity of electricity passed through the electrolytic solution or melt".
The reduction of one mol of Cu2+ to Cu can be represented as:
Cu2++ 2e-®Cu
Since, in this reaction there are two moles of electrons involved, so the amount of charge required is 2F.
(b) The cell reaction can be represented as:
Mg(s) + Cu2+(aq.) ® Mg2+(aq.) + Cu(s)
Given,Ecello= +2.71 VT = 298 KAccording to the Nernst equation:E = Ecello-0.05912log[Mg2+]Cu2+ = 2.71 - 0.05912log0.10.01
=2.71-0.0295 log 10 = 2.71 - 0.0295=2.6805 V
