RBSE Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

RBSE Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

Rajasthan Board RBSE Class 12 Chemistry Chapter 3 Electrochemistry

RBSE Class 12 Chemistry Chapter 3 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 3 Multiple Choice Questions

Question 1.
Which of the following is not a conductor?
(a) Cu_(metal)
(b) NaCl_(aq)
(c) NaCl_(molten)
(d) NaCl_(solid)

Question 2.
If conductance and conductivity of a cell are equivalent then cell constant will be :
(a) 1
(b) 0
(c) 10
(d) 1000

Question 3.
The unit of cell constant is
(a) ohm^-1cm^-1
(b) cm
(c) ohm^-1 cm
(d) cm^-1

Question 4.
The unit of conductivity (specific conductance) is:
(a) ohm^-1
(b) ohm^-1 cm^-1
(c) ohm^-1 cm² equiv^-1
(d) ohm^-1 cm²

Question 5.
If redox reaction takes place in cell then electromotive force (e.m.f.) of cell will be:
(a) positive
(b) negative
(c) zero
(d) one

Question 6.
Which statement will be correct for the cell made up of zinc and copper on the basis of electro chemical series?
(a) Zinc will act as cathode and copper as anode.
(b) Zinc will act as anode and copper as cathode.
(c) The flow of electrons takes place from copper to zinc.
(d) Copper electrode dissolves and zinc is deposited at zinc electrode.

Question 7.
How many coulomb charge will required for oxidation of one mole H₂O into O₂
(a) 1.93 x 10⁵ C
(b) 9.65 x 10⁴ C
(c) 6.023 x 10²³ C
(d) 4.825 x 10⁴ C

Question 8.
Which layer is electroplated at iron sheet?
(a) C
(b) Cu
(c) Zn
(d) Ni

Question 9.
Which of the following mixture represents rusting of iron?
(a) FeO and Fe(OH)₃
(b) FeO and Fe(OH)₂
(c) Fe₂O₃ and Fe(OH)₃
(d) Fe₃O₄ and Fe(OH)₂

Question 10.
When lead storage cell discharge, then
(a) SO₂ evolves
(b) Pb SO₄ destroys
(c) Pb forms
(d) H₂SO₄ destroys
Answers:
1. (d) 2. (a) 3. (d) 4. (b) 5. (a) 6. (b) 7. (b) 8. (c) 9. (c) 10.(d)

RBSE Class 12 Chemistry Chapter 3 Very Short Answer Type Questions

Question 1.
Can you store copper sulphate solution in a zinc pot?
Answer:
Zinc is more reactive than copper. Hence it displaces copper from copper sulphate solution as follow:
Zn_(s) + CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)
Thus, zinc reacts with CuSO₄ solution. Hence we cannot store copper sulphate solution in a zinc pot. Alternatenely,
E°_{zn²⁺/Zn = 0.76 V}
E°_{Cu²⁺ /Cu = 0.34 V}
To check whether the reaction will take place or not, we have to find out EMF of this cell
Zn_(s) + CuSO_{4 (aq)} → ZnSO_{4 (aq)} + Cu_(s)
Cell Zn/Zn²⁺ || Cu²⁺/Cu
E°_Cell = E°_(Cathode) – E°_(Anode)
= 0.34V – ( -0.76V)
E°_Cell = + 1.10 V
As E°_Cell is positive, so the reaction will take place. Hence we cannot store copper sulphate in a zinc pot.

Question 2.
Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Answer:
For the oxidation of ferrous ion, following reaction will take place
Fe²⁺ → Fe³⁺ + e^– E°_ox  = – 0.77 V
Only those sulstances can oxidise ferrous ion into ferric ion, which are stronger oxidising agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reeaction is positive. This is so for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series e.g Br₂, Cl₂, and F₂.

Question 3.
Why does the conductivity of solution decreases with dilution ?
Answer:
Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Hence the conductivity decreases.

Question 4.
Suggest a list of metals that are extracted electrolytically.
Answer:
Sodium, calcium, magnesium and aluminium.

Question 5.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methane and Methanol.

Question 6.
Arrange the following metals in the order in · which they displace each other from the solution of their salts. Ag, Al, Cu, Fe, Mg and Zn
Answer:
Mg, Al, Zn, Fe, Cu, Ag.

RBSE Class 12 Chemistry Chapter 3 Short Answer Type Questions

Question 1.
How would you determine the standard electrode potential of Mg²⁺/Mg?
Answer:
For determing the standard electrode potential of Mg²⁺/Mg, we have to set up a cell in which one electrode is Mg/Mg SO₄ (1 M), can be made by dipping a magnesium wire in 1 M MgSO₄ solution and the other one is hydrogen electrode Pt, H₂ (1 atm)/H⁺ (1 M). Now measure the e.m.f. of the cell. Here we have to note also the direction of deflection in the voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence the cell may be represented as,
Mg/Mg²⁺ (1 M) || H⁺ (1 M)/H₂(1 atm), Pt

Question 2.
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Answer:
For, pH = 10
pH = log [H⁺]
[H⁺] = 10^-pH
[H⁺] = 10^-10 mol L^-1
For hydrogen electrode,

Question 3.
Calculate the e.m.f. of the cell in which the following reaction takes place :
Ni_(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag (s) Givent that E°_Cell = 1.05 V
Answer:
Given that

= 1.05V – 0.134V = 0.916 V

Question 4.
The cell in which the following reaction occurs:
2 Fe³⁺_(aq) + 2I^–_(aq) → 2 Fe²⁺_(aq) + I_2(s) has E°_Cell = 0.236 V at 298 K. Calculate the standard Gibb’s energy and the equilibrium constant for the cell reaction.
Answer:
2 Fe³⁺ + 2e^– → 2 Fe ²⁺
2I^– → I₂ + 2e^–
Hence for the given cell reaction n = 2
∆_rG^o = – nFE°_Cell
= – 2 x 96500 x 0.236J
= – 45.55 kJ mol^-1
∆_rG^o = – 2.303 RT log Kc


= 7.983
K_c = Antilog (7.983)
= 9.616x 10⁷

Question 5.
Suggest a way to determine the ∧°_m( HCl ) – ∧°_m( NaCl )
Answer:
At infinite dilution, the limiting molar concentration of water ∧°_m( H₂O ) can be calculated by knowing the molar concentration of sodium hydroxide, hydrochloric acid and sodium chloride at infinite dilution.
∧°_m(H2O) = ∧°_m(HCl) + ∧°_m(NaOH) – ∧°_m(NaCl)

Question 6.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissoiation and dissociation constant. Given ∧°_m(H⁺) = 349.6 S cm² mol^-1 and ∧°_m(HCOO^–) = 546 S cm² mol^-1?
Answer:
∧°_m (HCOOH^–)= ∧°_m (H⁺) + ∧°_m (HCOO^–)
= 349.6 + 54.6
= 404.2 S cm² mol^-1

Question 7.
Consider the reaction
Cr₂O₇^2- + 14H⁺ + 6e^– →  2Cr³⁺ + 7H₂O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇^2-
Answer:
According to equation for the reduction of one mole of Cr₂O₇^2- , 6 mole of electrons are needed. Hence the amount of electricity require = 6F
= 6 x 96500 C = 579000 C

Question 8.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during
recharging.
Answer:
Lead storage battery can be recharged by passing electric current of suitable patential in opposite direction in which lead is again deposited at anode and PbO₂ is deposited at cathode. The density of electrolyle i.e., H₂SO₄ again increases. The following cell reaction takes place during charging :
At Anode : Pb_(s) + SO₄^-2_(aq) → PbSO_4(s) + 2e^–
At Cathode : PbO_2(s) + SO^-2_(aq) + 4H⁺_(aq) + 2e^– → PbSO_4(s) + 2H₂O_(l)
Overall reaction : Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)

Question 9.
Given the stardard electrode potentials
K⁺/K = – 2.93 V, Ag⁺/Ag = 0.80 V
Hg₂²⁺/Hg = 0.79V, Mg²⁺/Mg = – 237V, Cr²⁺/Cr = – 0.74 V
Arrange these metals in their increasing order of reducing power.
Answer:
Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus increasing order of reducing power will be Ag < Hg < Cr < Mg

Question 10.
Calculate the standard cell polentials of galvenic cells in which the following reactions take place :
(i) 2Cr_(s) + 3Cd²⁺_(aq) → 2Cr³⁺_(aq) + 3Cd_(s)
(ii) Fe²⁺_(aq) + Ag⁺_(aq) → Fe³⁺_(aq) + Ag_(s)
Given
E°_Cr³⁺/Cr = – 0.74 V, E°_Cd²⁺/Cd = 0.40 V
E°_Ag⁺/Ag = 0.80 V, E°_{Fe³⁺/Fe²⁺} = 0.77 V
Also calculate ArGo and equlibrium constants for the reactions.
Answer:
(1) 2Cr _(s) + 3Cd_(aq)²⁺ → 2Cr³ + 3Cd_(s)
E°_cell = E°_Cathode – E°_Anode
= E°_(Cd²⁺cd) – E°_(Cr³⁺/Cr)
= – 0.40V – (-0.74V)
= – 0.40 V + 0.74V
= + 0.34V
∆G° = -nFE°_Cell
= – 6 mol x 96500 C mol^-1 x 0.34 V
= – 196860 CV mol^-1
= – 196.860 J mol^-1
∆G° = – 2.303RT log K_c
– 196.860 kJ = – 2.303 x 8.314 x 298 x logK_c

log K_c = 34.50/4
K_c= Antilog 34.50/4
= 3.173 x 10³⁴
Hence Gibb’s energy of cell (∆G°)= – 196.86 kJ mol
Equilibrium constant (K_c) = 3.173 x 10³⁴
(ii) Fe²⁺_(aq) + Ag⁺_(aq) → Fe³⁺_(aq) + Ag_(s)
E°_Cell = E°_(Cathode) – E°_(Anode)
= E°_{(Ag⁺/ Ag)} – E°_{(Fe³⁺, Fe)}
= + 0.80 V – 0.77V
= 0.03V
∆G° = – n F E°_Cell
= – 1 mol^-1 x 96500 C mol^-1 x 0.03
= – 2895 J mol^-1
= – 2.895 kJ mol^-1
∆G° = – 2.303 RT log K_c
– 2895 = – 2.303 RT log K_c
– 2895 = – 2.303 x 8314 x 298 x logK_c

logK_c = 0.5074
K_c = Antilog 0.5074
K_c = 3.22
Gibb’s energy of cell = 2.895 kJ mol^-1
Equilibrium constant of cell = 3.22

RBSE Class 12 Chemistry Chapter 3 Long Answer Type Questions

Question 1.
Explain how rusting of iron is considered as setting up of an electrochemical cell.
Answer:
In rainy season and in highly humid atmosphere the iron surface has a layer of water. This water layer dissolves acidic oxides of the air like CO₂, SO₂ etc. to form acids which dissociate to give H⁺ ions.
H₂O + CO₂ → H₂CO₃ ⇌ 2H₊ + CO₃²⁺
In the presence of H⁺ ions, iron starts losing electrons at some spot to form ferrous ions, i.e., its oxidiation takes place. Here it is anode.
Fe_(s) → Fe²⁺_(aq) + 2e^–_(anode)
Now the released electrons move through the metal and reach to another spot where H⁺ ions and the dissolved oxygen take up these electrons and reduction reaction takes place. Here it is cathode.
O_2(g) + 4H⁺_(aq) + 4e^– +2H₂O_(l)(cathode)
The overall reaction is
2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)
Thus in this way an electrochemical cell is set up on the surface. This ferrous ion further oxidised by the atmospheric oxygen to ferric ions which combine with water molecules to from hydrated ferric oxide,
Fe₂O₃ x H₂O, which is rust.

Question 2.
Depict the galvenic cell in which the reaction
Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s)
takes place. Further, show,
(i) Which of the electrodes is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:
The given chemical reaction is as follows
Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s)
Cell can be represented as,
Zn_(s)/Zn²⁺_(aq) || Ag⁺_(aq)/Ag_(s)
(i) The electrode at which oxidation takes place, behaves as anode and it is negatively charged. Hence zinc electrode is negatively charged.
(ii) In the cell, the carriers of the current are electrons. and the current will flow from silver to copper in the external circuit.
(iii) The reactions on each electrode are as follows.
At anode : Zn_(s) → Zn²⁺_(aq) + 2e^–
At cathode : Ag⁺_(aq) + e^– → Ag_(s)
It can be represented by the following diagram

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