SikhLo - Current Affairs in Hindi, Daily current affairs, Current affairs

 Bihar Board Class 12 Chemistry Solutions Chapter 3 Electrochemistry

BSEB Bihar Board Class 12 Chemistry Solutions Chapter 3 Electrochemistry

Bihar Board Class 12 Chemistry Electrochemistry Intext Questions and Answers

Question 1.
How would you determine the standard electrode potential of the system Mg2+/Mg?
Answer:
The electrode of Mg dipping in the solution of its own ions Mg2+ is connected to Standard Hydrogen electrode (SHE) through a salt bridge internally and through voltmeter in the outer circuit to complete the cell

Cell reaction is Mg (s) + 2H+ (aq) → Mg2+ (aq) + H2 M. The e.m.f. of the ceil is determined which is 2.36 V.
Since SHE acts as cathod–
∴ According to conventiion, the standar electrode pen Mg2+/Mg system is – 2.36 V.

Question 2.
Can you store copper sulphate solution.
Answer:
In this problem, we want to see whether the tone takes place or not?
CuSO4 + Zn → ZnSO4 + Cu i.e.,
Zn (s) + Cu2+ → Zn2+ + Cu (s)

By convention, the cell maybe represented by
Zn (s) | Zn2+|| Cu2+ Cu (s)
EZn2+/Zn= -0.76 V
ECu2+/Cu = +0.34 V
E0Cell = E0R.H.S – E0L.H.S
= +0.34-(-0.76) = 1.1 V
Since the e.m.f. comes out to be positive, it implies that the reaction between CuSO4 and zinc will take place.
Hence CuSO4 solution cannot be stored in a zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Fe2+ → Fe3+ + e– oxidation
12 cl2 (g) + e– → Cl– ] reduction
Substances like F2, Cl2, Br2 which have more values of standard reduction potentials than that of Fe3+ [0.77V] will be able to oxidize Fe2+ ions to Fe3+ ions.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution where pH is 10.
Answer:
pH = 10
∴ [H+] – 1 × 10-10
H+ + e– → H2(g)
Applying Nemst’s equation, we get

EH+/H2 = E∘H+/H2−0.05911log1[H+]
E = 0 + 0.0591 log [H+] [E0 of NHE = 0]
= 0.0591 × log 10-10 [log 10 = 1
= 0.0591 (-10) log 10
EH+/H2 = -0.591 v

Question 5.
Calculate the emf of the cell in which the following reaction takes place. Given Eθcell = 1.05 V.
Ni (s) + 2 Ag+ (0.002 M) → Ni2+ (0.160 M) + 2 Ag (s)
Answer:
The probable cell is

Ecell = E0cell – 0.0591nlog[Ni2+][Ag]2[Ni][Ag+]2
Here n = 2
Ecell = E0cell – 0.05912log[Ni2+][Ag+]2
[ ∵ [Ni] = 1,[Ag] = 1]
Ecell = 1.05−0.05912log0.16(.002)2
Ecell = 0.91V.

Question 6.
The cell in which the following reaction occurs 2 Fe3+ (aq) + 2I– (aq) → 2 Fe2+ + I2 (s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer:
ΔrGθ = – nFEe = -2 × 96500 × 0.236 × 103 kj mol-1
= – 45.6 kj mol-1.
E°cell = 0.05912logKc
∴ 0.236×2.0591 = log Kc
∴ Kc= antilog 7.9864 = 9.7 × 107.

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
The conductivity of an electrolyte falls with dilution because the number of current-carrying particles, i.e., ions present per cm3 of the solution becomes less and less on dilution.

Question 8.
Suggest a way to determine the Am of water.
Answer:
Water is a very weak electrolyte. The value of Am of water can be determined using Kohlrausch’s law.
A0m(H2O) = λ∘m(H+)+λ∘m(OH−)
= [λ0m(H+)]+[λ0m(OH–)]
= A0m(HCl) + A0m(NaOH) – A0m(NaCl)

Question 9.
The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2mol-1. Calculate its degree of dissociation and dissociation constant. Given λ0H+ = 349.6 S cm2 mol-1 and λ0(HCOO–)= 54.6 S cm2mol-1.
Answer:
Degree of dissociation (a) of methanoic aid =ΔcΔ∘=ΔCλ∘H++λ0HCOO−
= 46.1349.6+54.6=46.1404.2 = 0.114

Dissociation constant of methanoic acid
= Cα21−α
= 0.025×(0.114)21−0.114
= 0.025×(0.114)20.886
= 0.025×0.0129960.886
= 3.67 × 10-4 mol L-1.

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire?
Answer:
Q = I × t = 0.5 × 2 × 60 × 60 C
= 3600 C
Charge present on 1 electron = 1.6023 × 10-19 C
∴ No. of electrons flowing = 36001.6023×10−19
= 2.25 × 1022
or No. of Moles of electrons = 2.25×10226.023×1023 = 3.73 × 10-2.

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Metals like Li, Na, K, Mg, Ca which are strong reducing agents themselves can’t be extracted by reducing their salts with other reducing agents.
∴ They are extracted from their molten solutions by electrolysis.

Question 12.
Consider the reaction:
Cr2O2-7 + 14H+ + 6e– → 2 Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O2-7?
Answer:
For reducing 1 mole of Cr2O2-7 ions 6 moles of electrons are required. For 1 mole of electrons ‘1F’ or 96,500 C of electricity is required.
∴ For passing 6 moles of electrons or 6F of electricity is required = 6 × 96,500 C = 579,000 C.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
During recharging the cell is operated like an electrolytic cell, i.e., now electrical energy is supplied to it from an external source. The electrode reactions are the reverse of those that occur during discharge.

Such operation is possible because PbSO4 formed during discharge is a solid and sticks to the electrodes. It is, therefore, in a position to either receive or give up electrons during electrolysis.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Two other substances which can he used as materials in fuel cells other than hydrogen are CO, methane (CH4) and methanol (CH3OH).

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer:
The rusting of iron in the presence of water and air is essentially an electrochemical phenomenon. A series oI tiny electrochemical cells are set upon the surface of iron.
Anode: 2 Fe (s) → 2Fe2+ + 4e– E° Fe2+/Fe = -0.44 V
Electrons are lost at anode Here Fe is oxidised to Fe2+. Electrons thus lost at anodic spot move through the metal and reduce oxygen in the presence of H+ ions.

Cathode: O2 +4H++4e– → 2H2O(l); E°H+|O2|H2O
The overall reaction is
2 Fe (s) + O2 (g) + 4H+ (aq) → 2Fe2+ + 2H2O (l); E°cell = 1.67 V
Atmospheric Oxidation takes place when Fe2+ ions are oxidized to Fe2O3 by oxygen of air
2 Fe2+ (aq) + 2 H2O (l) + 12O2(g) → Fe2O3 (s) + 4H+ (aq).

Bihar Board Class 12 Chemistry Electrochemistry Text Book Questions and Answers

Question 1.
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.
Answer:
The standard reduction of these metals are – 1.66 V, + 0.34 V, – 0.44 V, – 2.37 V and – 0.76 respectively.
The lower the reduction potential, more the tendency to lose electrons and thus more will be the displacement power. Thus the decreasing order of displacement power is Mg > Al > Zn > Fe > Cu.

Question 2.
Given the standard electrode potentials. K+/K = -2.93 V,Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V
Mg2+/Mg = – 2.37 V, Cr3+\Cr = – 0.74 V.
Arrange these metals in their increasing order of reducing power.
Answer:
The reducing power of a metal depends upon its tendency to lose Electrons. Thus lower the reduction potential, more the tendency to get oxidized and thus more will be the reducing power. Hence the increasing order of their reducing power is Al < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn (s) + 2 Ag+ (aq) → Zn2+ (aq) + 2 Ag (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.
Answer:
The cell reaction is Zn (s) + 2 Ag+ (aq) → Zn+2 (aq) + 2 Ag.
The reaction occuring at ANODE is
Zn (s) → Zn++ + 2e–] oxidation half reaction
The reaction occuring at CATHODE is
2 Ag+ (aq) + 2e– → 2 Ag (s)] reduction half reaction .

The complete cell reaction is
Zn (s) + 2 Ag+ (aq) → Zn2+ + 2 Ag (s)
The representation of the cell will be
Zn (s) |Zn++ (aq) || Ag+ (aq) | Ag (s)
N.D. Study Material Based On Depiction of the Galvanic Cell

Electrons flow from anode to cathode in the outer circuit. Conventional current flow against the flow of electrons from cathode to anode. Anode is given the negative sign and cathode is given the positive.

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) 2 Cr (s) + 3 Cd2+ (aq) → 2 Cr3+ (aq) + 3 Cd
(ii) Fe2+ (aq) + Ag+ (aq) →Fe3+ (aq) + Ag (s)
Calculate the ΔrGθ and equilibrium constant of the reactions.
Answer:
The galvanic cell may be represented as Cr (s) | Cr3+ (aq) || Cd2+ (aq) | Cd (s)
Reaction occuring at anode
Cr (s) → Cr3+ + 3e–] × 2
oxidation half reaction
Reaction occuring at cathode
Cd2+ + 2e– → Cd (s) ] × 3
reduction half reaction Adding the two half reactions;
cell reaction is
2 Cr (s) + 3 Cd2+ (aq) → 2 Cr3+ (aq) + 3Cd (s)
E0cell = E0Red[R-H.S.]-E°Red[L.H.S.]
E0Cd2+,Cd – E0Cr3+,Cr
= – 0.40 – (- 0.74) = + 0.34 VOLT.

ΔrGθ = -nF E0cell
Here n = 6 F= 96,500 C
E0cell = +0.34 V
ΔrGθ = -6 × 96500 × 0.34
= -1.968 ×105J [1C × 1 V = 1J]

Calculation Of K
ΔrGθ = – 2.303 RT log K 1
= – 2.303 × 8.314 × 298 log K
[∵R = 8.314 JK-1 Mol-1 T = 298 K]
Now ΔrGθ = -1.968 × 105 J(Calculated above)
∴ -1.968 × 105 = – 2.303 × 8.314 × 298 log K
log K = 1.968×1052.303×8.314×298
K = antilog 34.501 = 3.17 x 1034

(ii) The galvanic cell may be represented by
Fe2+ (aq) |Fe3+ (aq) || Ag+ (aq) | Ag (s)
The reaction at anode
Fe2+→ Fe3+ + e– ] Oxidation half reaction

The reaction at cathode
Ag+ (aq) + e– → Ag (s) ] Reduction half reaction

E°cell = E0Red[R.H.S.]-E0Red[L.H.S.]
= E°Ag+,Ag – E°Fe2+,Fe3+
= (0.80-0.77) = 0.03 V
ΔrGθ= – n F E°cell [n = 1, F = 96,500 C]
E°cell = -(1) (96500) (0.03)
= – 2895 J mol-1

ΔrGθ = – 2895 J = -2.895 KJ mol-1
ΔrGθ = – 2.303 RT log K
log K = –ΔrGθ2.303RT
= – −28952.303×8.314×298
= 0.5074
K = antilog 0.5074 = 3.217

Question 5.
Write the Nemst equation and emf of the following cells at 298 K:
(i) Mg (s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M)| Cu (s)
(ii) Fe (s) | Fe2+ (0.001 M) |] H+ (1M) | H2 (g) (1 bar) | Pt (s)
(iii) Sn (s) | Sn2+ (0.050 M) ||H+ (0.020 M) | H2 (g) (1 bar) | Pt (s)
(iv) Pt (s) | Br2 (l) | Br (0.010 M) ||H+ (0.030 M) | H2 (g) (1 bar) | Pt (s).
Answer:
Mg (s) | Mg2+ (.001 M) || Cu2+ (0.0001 M) | Cu (s)
Reduction reaction are
Mg+2 + 2e– → Mg(s)
Cu+2 + 2e– → Cu (s)
Cell reaction is
Mg (s) + Cu++ → Mg++ + Cu (s)

Ecell = E°cell – RTnFln[Mg+2][Cu][Mg][Cu++]
Ecell = E°cell – RTnFln[Mg2+][Cu2+]
Since [Cu] = 1
[Mg] = 1

E°Cu++,Cu – E°Mg+ + ,Mg
= – 0.05912log0.0010.0001 [ n= 2]
= [0.34 -(-2.36)]
= 2.7 – 0.0295 log 10 [log 10 = 1]
Ecell = (2.7-0.0295) V = 2.68 V

(ii) Fe (s) | Fe2+ (0.001 M) || H+ (1 M), H2 (g) (1 bar) | Pt (s)
Cen reaction is
Fe (s) +2n → Fe2+ + H2 (g)
Ecell = E°cell – RTnFln[Sn2+][H+]2 [Sn(s)= 1]
= [ E° H+,H2 – E° Sn+,Sn] – 0.05912log0.05(0.02)2 n = 2 [Sn2+] = 0.05 M
Ecell = [0- (-0.14)]- 0.05912 log 125
= 0.14 – 0.0295 × 2.0969 [H+] = 0.02 M
= 0.14 – 0.06 [ log 125 = 2.0969]
Ecell = 0.08 V

(iv) Pt (s) | Br2 (I) | Br– (0.01 M) ||H+ (0.03 M) | H2 (g) (1 bar) | Pt (s)
Cell reaction is:
2 Br– + 2H+ → Br2 + H2
Ecell = E°cell – 0.05912log1[ Br¯]2[H+]2
Ecell = (0 – 1.08) – 0.05912log1(0.01)2(0.03)2
= – 1.08 – 0.05912(7.0457)
=-1.08 – 0.208 = -1.288 V
Thus, oxidation will occur at the hydrogen electrode and reduction at the Br2 electrode
In that case Ecell = 1.288 V

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:
Zn (s) + Ag2O (s) + H2O (l) → Zn2+ (aq) + 2 Ag (s) + 2OH– (aq) Determine ΔrGθ and Eθ for the reaction.
Answer:
Given Zn (s) → Zn2+ + 2e–; E° = 0.76 V ’
Ag2O + H2O + 2e– → 2 Ag + 2OH–;E° = 0.344 V
E°cell = E°Ag°2 ,Ag + E°Zn,Zn2+ )
= 0.344 + 0.76 = 1.104 V ‘
ΔrG° = -nF E°cell = – 2 × 96,500 × 1.104 J ,
= -2.13×105J

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
Conductivity of specific conductance (K, α : Kappa: Greek , word) is the reciprocal of resistivity. .
It is the conductance of a material which is 1 metre in length and; having an area of cross-section of 1m2. v
Its unit is Siemen (S) m-1.
1S cm-1 = 100 Sm-1
Siemen or 1 S = 1 ohm-1 or mho or Ω-1 If we take an electrolytical solution in a 1 cm3 of a vessel having: electrodes 1 cm apart, its conductance is also called specific conductance or conductivity.
Unit of K is also ohm-1 cm-1.

Molar Conductivity-It is the conducting power of all ions producing by dissolving one mole of an electrolyte in solution. It is denoted by Am (lambda). Molar conductivity is related to specific conductance/conductivity (x) as ∧m = K/M. where M is the molar concentration. If M is in the units of Molarity, i. e., moles per litre (mol L-1), ∧m may be expressed as ∧m = κ×1000M
Unit of Molar conductivity
Molar conductivity =’ S m2 mol-1

Specific conductance or conductivity decreases on dilution as the * number of ions per cm3 of the solution decreases on decreasing the concentration or increasing the dilution.
The variation of Molar conductance or molar conductivity with concentration can be explained on the basis of conducting ability of ions for weak and strong electrolytes.

(i) Conductivity behaviour of weak electrolytes-The variation of Am with dilution can be explained on the basis of number of ions in solution. The number of ions furnished by an electrolyte depend upon the degree of dissociation with dilution. With Increase in dilution, the degree of dissociation increases and as a result molar conductivity increases. The limiting value of molar conductance/conductivity (Δm∞) corresponds to degree of dissociation equal to one, i.e., the whole of the I electrolyte dissociates.

Thus the degree of dissociation can be calculated at any concentration as α = ΔcmΔ∞m
where a is the degree of dissociation, Δmc is the molar conductance at infinite dilution.

Molar Conductivity of Strong Electrolytes In the case of strong electrolytes there is no increase in the number of ions with dilution because strong electrolytes are completely ionised in solution at all concentrations. However, in concentrated solution of strong electrolytes, there are strong forties of attraction between the ions of opposite charges called inter-ionic forces. Due to these inter-ionic forces, the conducting ability of the ions is less in concentrated solutions. With dilution, the ions become far apart from one another and interionic forces decrease.

As a result molar conductivity increases with dilution. When the concentration of the solution becomes very very low, the interionic attraction becomes negligible and the molar conductivity approaches the limiting value called Molar Conductivity At Infinite Dilution. It is characteristic of each electrolyte ∧m and Δm∞ for strong electrolytes are related to each other by ∧m = Δm∞ – bc√;
b is a constant which depends upon the viscosity and dielectric constant of the solvent; c is the concentration of the solution.

Question 8.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.
Answer:
K (conductivity) Molarity = 0.0248 S cm-1 = 0.20 M
Molarity = 0.20 M
∧m = κ×1000 Molarity 
= 0.0248 S cm−1×1000 cm3 L−10.20 mol L−1
= 124 S cm-1 mol-1
= 124 S cm2 mol-1
∴ Molar conductivity = 124 S cm2 mol-1.

Question 9.
The resistance of a conductivity cell containing 0.001 KCl solution at 298 K is 1500 Ω. What is the cell constant if conductive of 0.001 M KC1 solution at 298 K is 0.146 × 10-3 S cm-1.
Answer:
Resistance of the conductivity cell = 1500 Ω
∴ Conductance = 11500 ohm-1 or S
Conductivity (K) of solution = 0.146 × 10-3 S cm-1
∴ Cell Constant(la) =  Specific conductance  Observed conductance 
= 0.146×10−311500
= 0.146 × 10-3 × 1500 cm-1
Cell constant– = 0.219 cm-1

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Am for all concentrations and draw a plot between ∧m and c12. Find the value of ∧m0
Answer:


∧° = intercept on Am axis = 124.0 S cm2 mol-1
on extrapolation to zero concentration

Question 11.
Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if ∧m0 for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
Answer:
(K) Conductivity of acetic acid = 7.896 × 10-5 S cm-1
for acetic acid = 390.5 S cm2 mol-1

Δmc = κ×1000 Molarity 
= 7.896×10−5×10000.00241
= 32.76 S cm2 mol-1

α = degree of dissociation = ΛcmΛ∘m=32.76390.5 = 8.4 × 10-2
Dissociation constant of acetic acid Ka = Cα21−α
= 0.00241×(8.4×10−2)21−0.084
= 1.86× 10-5.

Question 12.
How much charge is required for the following reductions
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO4– toMn2+.
Answer:
(i) The electrode reaction is Al3+ + 3 e– → Al
∴ Quantity of charge required for reduction of 1 Mole of Al3+
= 3 × F = 3 × 96500 C
= 289500 C

(ii) The electrode reaction is
Cu2+ + 2 e– → Cu
Quantity of charge required for reduction of 1 Mole of Cu2+
= 2 × F = 2 × 96,500 C = 1,93,000 C

(iii) The electrode reaction is
MnO4– → Mn2+
i. e., Mn7+ + 5 e– → Mn2+
Quantity of charge required, = 5F
= 5 × 96,500 C = 482500 C.

Question 13.
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2
(ii) 40.0 g of A1 from molten Al2O3.
Answer:
(i) Ca2+ + 2 e–→ Ca
Thus 1 mol of Ca (= 40 g) require electrictiy = 2F
∴ 20 g of Ca will require = 1 F

(ii) Al3+ + 3 e– → Al
Thus 1 mol of Al (= 27 g) require electricity = 3F
∴ 40 g of A1 will require electricity = 327 × 40 = 4.4 F.

Question 14.
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2?
(ii) 1 mol of FeO to Fe2O3?
Answer:
(i) The electrode reaction for 1 mol of H2O is
H2O →H2+12 O2
i.e., O2- → 12O2 + 2 e
∴ Quantity of electricity required = 2F
= 2 ×96,500 = 1,93,000 0

(ii) The electrode reaction for 1 mol of FeO is
FeO → 12Fe2O3
i.e., Fe2+ → Fe3+ + e–
Quantity of electricity required = F = 96,500

Question 15.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
Atomic mass of N,i = 58.7
time = 20 min = 20 × 60 = 1200 sec.
Quantity of electricity passed= I × t = 5 × 1200 = 6000 C
Ni (NO3)2 will yield Ni2+ ions in the solution .

Thus, 2F, i.e., 2 × 96,500 C deposit Ni = 1 Mole = 58.7 g
∴6000 C will deposit Ni = 58.72×96,500 × 6,000 g = 1.825 g
∴Mass of Ni deposited at cathode = 1.825 g

Question 16.
Three electrolytic cells A, B, C containing solution of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and of zinc were deposited?
Answer:

∴ Zn deposited = 65.32×96500×1295.6 = 0.438 g.

Question 17.
Using the standard electrode potentials given in the Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+ (aq) and I– (aq)
(ii) Ag+ (aq) and Cu (s)
(iii) Fe3+ (aq) and Br– (aq)
(iv) Ag (s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq).
Answer:
Given standard electrode potentials:
E°12 I2,I– = 0.54 V E°Cu2+,Cu = 0.34
E°12 Br2,Br– = 1.09 V E°Ag+,Ag = +0.80 V
E°Fe3+,Fe2+ = 0.77 V
A Reaction is feasible if EMF of the cell reaction is positive.
(i) Fe3+(aq)+I–(aq) → Fe2++12+I2

E°cell = E°Fe3+,Fe2+ – E°12I2,I–
The reation is feasible = 0.77 – 0.54 = 0.23 V,

(ii) 2 Ag+ (aq) + Cu (s) →2 Ag (s) + Cu2+
E°cell = E°Ag+ ,Ag – E°Cu2+ ,Cu
[as the cell is Cu | Cu2+ || Ag+|Ag]
= 0.80 – 0.34 = 0.46 V

(iii) Fe3+ (aq) + Br– (aq) → Fe2+ (aq) +12 Br2
E°cell = 0.77-1.09 = 0.32 V
The reaction is not feasible.

(iv) Ag (s) + Fe3+ (aq) → Ag+ (aq) + Fe2+ (aq)
E°cell = 0.77 – 0.80 = -0.03 V
The reaction is not feasible.

(v) 12 Br2(aq) + Fe2+ → Br–(aq)+Fe3+(aq)
E°cell = 1.09-0.77 = 0.32 V
The reaction is feasible.

Question 18.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Answer:
(i) Electrolysis of aqueous solution of AgNO3 with silver electrodes
AgNO3(aq) → Ag+(aq) + NO3– (aq)
H2O ⇌ H+ + OH–
At Cathode: Ag+ ions have lower discharge potential than H+ ions.

Hence Ag+ ions will be deposited in preference to H+ ions.
Ag+ + e–→ Ag
At Anode: As Ag anode is attacked by NO3, Ag of the electrode will dissolve to form Ag+ ions in the solution.
Ag(s) → Ag+ + e–

(ii) Electrolysis of aqueous solution of AgNO3 with platinum electrodes.
At Cathode: Same as above.
At Anode: As anode is not attackable, out of OH– ions and NO3– ions, OH– ions have lower discharge potential. Hence OH– ions will be discharged in preference to NO3– which then decompose to give out O2
OH– (aq) → OH + e–
4 OH → 2H2O (l) + O2(g)

(iii) Electrolysis of dil. H2SO4 with platinum electrodes
H2SO4(aq) → 2H+(aq) + SO4-2 (aq)
H2O ⇌ H+ + OH–
At Cathode: H+ + e → H
H + H →H2
At Anode: OH– →OH + e
4OH → 2H2O (l) + O2 (g)

(iv) Electrolysis of aq. solution of CuCl2 with platinum electrodes
CuCl2 (s) + aq.→Cu2+ (aq) + 2 Cl– (aq)
H2O ⇌ H+ + OH–
At Cathode: Cu2+ ions will be reduced in preference to H+ ions.
Cu2+ + 2e– → Cu (s)
At Anode: Cl– ions will be oxidized in preference to OH– ions.
Cl– → Cl + C;Cl + Cl→Cl2(g)
Thus Cu will be deposited at cathode and Cl2 will be liberated at anode.