Short Questions Answer

Chemistry : Electro Chemistry

Short Answer Questions

1. Define anode and cathode

i) Anode: The electrode at which the oxidation occurs is called the anode.

ii) It is negative in a galvanic cell.

i) Cathode: The electrode at which the reduction occurs is called the cathode.

ii) It is positive in a galvanic cell.

2. Why does conductivity of a solution decrease on dilution of the solution

On dilution of the electrolyte solution, the ions present in the unit dimension was decreased and hence the conductivity of a solution also decreases.

3. State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.

Kohlrausch Law:

At infinite dilution, the limiting molar conducutivity of an electrolyte is equel to the sum of the limiting molar conductivites of its constituent ions.

The molar conductance at infinite dilution for weak electrolytes can be calculated using Kohlrausch’s law.

Molar conductivity weak electrolyte :

The molar conductance of CH₃COOH, can be calculated using the experimentally determind molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa.

Ʌ^o_HCl3COONa = λ^o_Na + λ_oCH3COO^-  …………. (1)

Ʌ^o_HCl = λ^o_H+ + λ^o_Cl^-   ---------------(2)

Ʌ^o_NaCl = λ^o_Na+ + λ^o_Cl^-   ………...(3)

Equation (1) + Equation (2) – Equation (3) gives,

(Ʌ^o_CH3COONa) + (Ʌ^o_HCl) – (Ʌ^o_NaCl) = λ^o_H+ + λ^o_CH3COO-

= Ʌ^o_CH3COOH

4. Describe the electrolysis of molten NaCl using inert electrodes

The products of electrolysis of molten NaCl is sodium metal and Cl₂ gas.

Na + e⁻ → Na

The anion Cl⁻ oxidized at the anode.

Cl → ½ Cl₂ + e⁻

5. State Faraday’s Laws of electrolysis

First law:

The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.

i.e., m α Q

m α lt or m= Z lt

Second law:

When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

m₁ ∝ Z₁

m₂ ∝ Z₂

m₁/Z₁ = m₂/Z₂

Z = Electrochemical equivalent

6. Describe the construction of Daniel cell. Write the cell reaction.

i) Daniel cell or a galvanic cell is an example of electro chemical cell.

ii) This overall reaction is made of the summation of oxidation half reaction and reduction half reaction.

iii) The oxidation half cell reaction occurring at the zinc electrode.

Zn(s) → Zn²⁺_(aq) + 2 e⁻

iv) The reduction half reaction occurring at the copper electrode. It receives the electrons from the zinc electrode when connected externally, to produce metallic copper according to the reaction as

Cu²⁺_(aq) + 2e⁻ → Cu_(s)

The over all reaction taking place in the cell is the redox reaction gives as

Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s)

v) The decrease in the energy which appears as the heat energy when a zinc rod is directly dipped into the ZnSO₄ solution, is converted into electrical energy. When the same reaction takes place indirectly in an electro chemical cell.

vi) When the cell is set up, electrons flow from zinc electrode through the wire to the copper cathode.

vii) As a result, zinc dissolves in the anode solution to form Zn²⁺ ions. The Cu²⁺ ions in the cathode half cell pick up electrons and are converted to Cu atoms on the cathode.

7. Why is anode in galvanic cell considered to be negative and cathode positive electrode?

• Oxidation occurs at anode, Electrons are liberated at anode and hence it is negative.

• Reduction occurs at cathode. The electrons are consumed at cathode and hence it is positive.

8. The conductivity of a 0.01M solution of a 1 :1 weak electrolyte at 298K is 1.5 ×10-4 S cm−1.

i) molar conductivity of the solution

ii) degree of dissociation and the dissociation constant of the weak electrolyte

Given that

λºcation = 248.2 S cm2 mol−1

λºcation = 51.8 S cm2 mol−1

Answer:

Given:

C = 0.01M

λ°_cation = 248.2 S cm² mol⁻¹

K = 1.5 × 10^-4 S cm⁻¹

λ°_anion = 51.8 S cm² mol⁻¹

(i) molar conductivity

Ʌ^o_m = [ k ×10⁻³ / M ] mol⁻¹m³

Ʌ^o _m= (1.5 × l0^-4 × l0⁻³ × l0² ) / ( 1 ×10^-2)

Ʌ°_m = 1.5 × 10⁻³ S m² mol⁻¹

(ii) Degree of dissociation

α = Ʌ^o_m / Ʌ^o_α

Ʌ^o_m = λ^o₊ + λ^o_-

Ʌ^o_∞ = λ^o_cation + λ^o_anion

= (248.2 + 51.8) S cm² mol⁻¹

= 300 S cm² mol⁻¹

Ʌ^o_∞ = 300 × 10^-4 S m² mol⁻¹

Ʌ^o_m = 1.5 × 10⁻³ Sm² mol⁻¹

α = Ʌ^o_m / Ʌ^o_∞ = (1.5 × 10⁻³ S m² mo1⁻¹) / (300 × 10^-4 S m² mol⁻¹)

α = 0.05

Dissociation constant

K_a = α²C / 1-α

= (0.05)²(0.01) / 1-0.05

= (25 × 10^-4 × 10^-2 ) / 95 × 10^-2

 = 0.26 × 10^-4

= 2.6 × 10^-5

Answer:

i) Molar conductivity = 1.5 × 10⁻³ S m² mol⁻¹

ii) Degree of dissociation = 0.05

Dissociation constant = 2.6 × 10^-5

9. Which of 0.1M HCl and 0.1 M KCl do you expect to have greater Λºm and why?

0.1 M HCl shows greater acidity

HCl dissociated as H⁺ and Cl⁻

KCl dissociated as K⁺ and Cl⁻

HCl releases H⁺.

The substance releases H⁺ is acid. Hence 0.1M HCl is more acidic than 0.1M KCl.

10. Arrange the following solutions in the decreasing order of specific conductance.

i) 0.01M KCl ii) 0.005M KCl iii) 0.1M KCl iv) 0.25 M KCl v) 0.5 M KCl

Answer:

i) 0.5 M KCl

ii) 0.25 M KCl

iii) 0.1 M KCl

iv) 0.01 M KCl

v) 0.005 M KCl

11. Why is AC current used instead of DC in measuring the electrolytic conductance?

When DC current is applied through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.

Hence AC current is used for this measurement to prevent electrolysis.

12. 0.1M NaCl solution is placed in two different cells having cell constant 0.5 and 0.25cm-1 respectively. Which of the two will have greater value of specific conductance.

Solution:

K = C × (ℓ/A)

Specific conductance is directly proportional to cell constant. Higher the cell constant higher will be the value of specific conductance. NaCl placed in cell having 0.5 cm⁻¹ shows greater value of specific conductance.

13. A current of 1.608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and the current efficiency is 100%.

Given: I = 1.608A;

t = 50 min = 50 × 60 = 3000s

 V = 500 mL; C = 0.5 M ; ɳ = 100%

Solution:

The number of Faradays of electricity passed through the CuSO₄ solution

⇒ Q = It

Q = 1.608 × 3000

Q = 4824 C

∴ Number of Faradays of electricity

= 4824 C / 96500 C = 0.5F

Electrolysis of CuSO₄

 Cu²⁺ (aq) + 2e⁻ → Cu(s)

The above equation shows that 2F electricity will deposit 1 mole of Cu²⁺ to Cu.

∴ 0.5F electricity will deposit ( 1 mol / 2F ) × 0.5 F

 = 0.025 mol

Initial number of moles of Cu²⁺ in 250 ml of solution = ( 0.5 / 1000 mL ) × 250 mL

 = 0.125 mol

∴ Concentration of Cu²⁺ = [ 0.1 mol / 250 mL ] × 1000 mL

 = 0.4M 

14. Can Fe3+ oxidises bromide to bromine under standard conditions?

Given: EFe 3+|Fe2+ = 0.771

EBr2|Br− = 1.09V.

Answer:

Required half cell reaction

 2Br⁻ → Br₂ + 2e⁻            (E°_ox) = −1.09V

2Fe³⁺ + 2e⁻  → 2Fe²⁺           (E°_red) = +0.771V

2Fe²⁺ + 2Br ⁻ → 2Fe²⁺ + Br₂ (E°_cell) = ?

(E°_cell) = (E⁰_ox) + (E°_red)

= −1.09 + 0.771

= −0.319 V

E°_cell is –ve; ∆G is +ve and the cell reaction is non spontaneous. Hence Fe³⁺ cannot oxidizes Br⁻ to Br₂.

15. Is it possible to store copper sulphate in an iron vessel for a long time?

Given : ECu 2+|Cu = 0.34 V and EFe 2+|Fe = −0.44V .

Answer:

Given: E^o_Cu2+_{| Cu} = 0.34 V and E°_Fe2+_|Fe = − 0.44V

(E°_ox)_{Fe | Fe}²⁺ = 0.44V and (E°_red)_Cu2+ _{| Cu} = 0.34V

The +ve emf values shows that iron will oxidise and copper will get reduced ie., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

16. Two metals M1 and M2 have reduction potential values of -xV and +yV respectively. Which will liberate H2 and H2SO4.

Answer:

Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having + xV, oxidation potential will liberate H₂ from H₂SO₄.

17. Reduction potential of two metals M and M2 are EºM12+|M1 = − 2.3V and E EºM22+|M2 = 0.2V  Predict which one is better for coating the surface of iron. Given : E Fe2+ |Fe = −0.44V

Answer:

Oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

18. Calculate the standard emf of the cell: Cd | Cd 2+ || Cu 2+ | Cu and determine the cell reaction. The standard reduction potentials of Cu2+ | Cu and Cd2+ | Cd are 0.34V and -0.40 volts respectively. Predict the feasibility of the cell reaction.

Answer:

Cell reactions:

Oxidation at anode : Cd(s) → Cd²⁺ (aq) + 2e⁻

Reduction at cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)

(E°_ox) _{Cd ∣ Cd}2+ = 0.4V

(E°_red) _{Cu2+  | Cu}= 0.34V

E°_cell = E°_R - E°_L

= 0.4 + 0.34

= 0.74V

emf = +ve ; ∆G = −ve

The reaction is feasible.

19. In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised at the anode and O2 at cathode. If 44.8 litre of H2 at 25oC and 1atm pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+ , how many grams of Cu deposited?

Oxidation at anode:

2H₂(g) + 4OH⁻ (aq) → 4H₂O (𝑙) + 4e⁻

1 mole of hydrogen gas produces 2 moles of electrons at 25 °C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres.

∴ no.of moles of hydrogen gas produced

= [1 mole / 22.4 litres] × 44.8 litres

= 2 moles of hydrogen

∴ 2 moles of hydrogen produces 4 moles of electron ie., 4F charge.

Q = It

 I = Q / t

= (4F) / (10 minutes)

= (4 × 96500 C ) / ( 10 × 60 s )

I = 643.33 A

Electro deposition of copper

Cu²⁺ (aq) + 2e⁻ → Cu(s)

2F charge is required to deposit

1 mole of copper ie., 63.5 g

If the entire current produced in the fuel cell ie., 4F is utilized for electrolysis, then 2×63.5 ie., 127.0 g copper will be deposited at cathode. 

20. The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Give : molar mass of Nickel and chromium are 58.74 and 52gm-1 respectively.

Answer:

Ni²⁺ (aq) + 2e⁻ → Ni(s)

Cr³⁺ (aq) + 3e⁻ → Cr (s)

The above reaction indicates that 2F charge is required to deposit 58.7g of Nickel from nickel nitrate and 3F charge is required to deposit 52 g of chromium.

Given that 2.935 gram of Nickel is deposited

∴ The amount of charge passed through the cell = [ 2F / 58.7g ] × 2.935g

= 0.1 F

∴ If 0.1 F charge is passed through chromium nitrate the amount of chromium deposited

= [52g / 3F] × 0.1 F = 1.733 g

21. A copper electrode is dipped in 0.1M copper sulphate solution at 25oC . Calculate the electrode potential of copper. [Given: ECu 2+|Cu = 0.34 V ].

Answer:

Solution:

[Cu²⁺] = 0.1M ;

E°_Cu2+ _{∣ Cu} = 0.34 V

E_cell = ?

Cell reaction is

Cu²⁺ (aq) + 2e⁻ → Cu(s)

E_cell = E°_cell – { 0.0591/n log ([Cu] / [Cu²⁺]) }

= 0.34 – {(0.0591/2) log (1/0.1)}

= 0.34 – 0.0296

E_cell = 0.31 V

22. For the cell Mg (s) | Mg2+ (aq) || Ag+ (aq) | Ag (s), calculate the equilibrium constant at  25oC and maximum work that can be obtained during operation of cell. Given : Eº Mg 2+ | Mg = −2.37V and EºAg + | Ag = 0.80V.

Answer:

Oxidation at anode

Mg → Mg²⁺ + 2e⁻         ------(1)

(E°_ox) = 2.37 V

Reduction at cathode

Ag⁺+ e⁻ → Ag           --------- (2)

(E°red) = 0.80 V

 E°_cell = (E°_ox)_anode + (E°_red)_cathode

= 2.37 + 0.80

= 3.17 V

Overall reaction

Eqn (1) + 2 × eqn (2) ⇒

Mg + 2Ag⁺ → Mg²⁺ + 2Ag

∆G° = −nFE°

= −2 × 96500 × 3.17

= −611.810 J

 ∆G°= −6.12 × 10⁵ J

∆G° = −2.303 RT log K_c

⇒ log K_c = [ −6.12 × 10⁵ ] / [2.303 × 8.314 × 298]

K_c = Antilog of (107.2)

23. 8.2 ×1012 litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 ×106 Cs−1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.

Answer:

Hydrolysis of water

At anode:

2H₂O → 4H⁺ + O₂ + 4e⁻    ----------(1)

At cathode:

2H₂O + 2e⁻ → H₂ + 2OH⁻      --------(2)

Overall reaction

6H₂O → 4H⁺ + 4OH⁻ + 2H₂+O₂

(or) Eqn (1) + (2) × 2 ⇒ 2H₂O → 2H₂ + O₂

∴ According to Faraday's law of electrolysis, to electrolyse two moles of water (36g = 36 mL of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = 4 × 96500 C.

∴ When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

= [ 4 × 96500C / 36 mL ] × 9 × 10¹² L

= [ (4 × 96500 × 9 × 10¹²) / (36 × 10⁻³)]  ×  C

= 96500 × 10¹⁵ C

∴ Given that in 1 second, 2 × 10⁶ C is generated therefore, the time required to generate

96500 × 10¹⁵ C is = [ 1S / 2×10⁶C ] × 96500 × 10¹⁵ C

 = 48250 × 10⁹ S

1 years = 365 days

= 365 × 24 hours

= 365 × 24 × 60 min

= 365 × 24 × 60 × 60 sec

∴ Number of years = [ 48250 × l0 ⁹ ] / [365 × 24 × 60 × 60]

= 1.5299 × 10⁶ years 

24. Derive an expression for Nernst equation

Nernst equation relates the cell potential and the concentration of the species involved in an electrochemical reaction.

Consider an electrochemical reaction is

xA + yB ⇌ lC + mD

The reaction quotient Q for the above reaction is given below

Q = [C]^𝑙 [D]^m  /  [A]^x[B]^y ……………(1)

∆G = ∆G° + RT InQ   …………… (2)

The Gibbs free energy can be related to the cell emf

∆G = −nFE_cell ;

∆G° = −nFE°_cell

Substitute ∆G° and Q in the eqn (2)

−nFE_cell = −nFE°_cell + RT ℓn { [C]^l[D]^m / [A]^x[B]^y }

Divide the whole eqn by (−nF)

⇒ E_cell = E°_cell – { 2.303 RT / nF   log. ( [C]^I[D]^m / [A]^x[B]^y ) }

This equation is called the Nernst equation.

25. Write a note on sacrificial protection.

It is a method to prevent the objects from corrosion. The object is covered with the protecting metals which is corrected more easily than the object. The metal corrods itself but saves the object.

Eg: Fe object is protected by Mg or Zn.

26. Explain the function of H2 - O2 fuel cell.

Fuel cell

The energy of combustion of fuels is directly converted into electrical energy is called the fuel cell. The general representation of a fuel cell is follows

Fuel ∣ Electrode ∣ Electrolyte ∣ Electrode ∣ Oxidant

Fuel - Hydrogen

Electrolyte - Aqueous KOH

Oxidant - Oxygen

Electrode - Porous graphite containing Ni

Ni and NiO serves as the inert electrodes.

Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.

Oxidation occurs at the anode:

2H₂(g) + 4OH⁻ (aq) → 4H₂O (ℓ) + 4e⁻

Reduction occurs at the cathode

O₂(g) + 2H₂O(ℓ) + 4e⁻ → 4OH⁻ (aq)

The overall reaction is

2H₂(g) + O₂ (g) → 2H₂O(ℓ)

27. Ionic conductance at infinite dilution of Al3+ and SO42- are 189 and 160 mho cm2 equiv-1. Calculate the equivalent and molar conductance of the electrolyte Al2 (SO4 )3 at infinite dilution.

Molar conductance

Ʌ°_m (Al₂(SO₄)₃) = 2λ°_Al3+ + 3λ°_SO42−

= (2 × 189) + (3 × 160)

Ʌ°_m = 858 mho cm² mol⁻¹

Equivalent conductance

 λ_∞ (Al₂SO₄)₃) = ( 1/3 λ_∞ Al³⁺ ) + ( 1/2 λ_∞ SO₄²⁻ )

= ( 1/3 × 189) + ( 1/2 × 160 )

= 63 + 80 = 143 mho cm² equiv⁻¹

EVALUATE YOURSELF:

1. Calculate the molar conductance of 0.01M aqueous KCl solution at 25°C. The specific conductance of KCl at 25°C is 14.114 × 10 ^-2 Sm⁻¹.

Given:

C = 0.01 M

k = 14.114 × 10^{− 2} Sm⁻¹

Ʌ_m = ?

Ʌ_m = (K × 10⁻³) / M

= ( 14.11 × 10^-2 × 10⁻³ ) / 0.01

= (14.11 × 10^-2 × 10 ⁻³ ) / 10^-2

Ʌ_m = 14.114 × 10⁻³ Sm²mol⁻¹

2. The resistance of 0.15N solution of an electrolyte is 50 Ω. The specific conductance of the solution is 2.4 Sm⁻¹. The resistance of 0.5 N solution of the same electrolyte measured using the same conductivity cell is 480 Ω. Find the equivalent conductivity of 0.5N solution of the electrolyte.

Given:

R₁ = 50 Ω

R₂ = 480 Ω

K₁ = 2.4 Sm⁻¹

K₂ = ?

N₁ = 0.15 N

N₂ = 0.5 N

Ʌ = [ K₂(Sm ⁻¹) × 10^{- 3}(gram equivalent)⁻¹ m³ ] / N

We know that

K = Cell constant / R

∴ K₂/ K₁ = R₁/R₂

K₂ = K₁ × [R₁/R₂]

= 24 Sm⁻¹ × (50Ω / 480Ω)

= 0.25 Sm⁻¹

= [ 0.25 × 10^{- 3} S (gram equivalent) ⁻¹m² ] / 0.5

Ʌ = 5 × 10^-4 Sm² gram equivalent⁻¹

3. The emf of the following cell at 25°C is equal to 0.34V. Calculate the reduction potential of copper electrode.

Pt(s) ∣ H2 (g, 1 atm) ∣ H⁺ (aq, 1M) ∣ ∣ Cu²⁺ (aq, 1M) I Cu(s)

E°_cell = (E°_ox)_anode + (E°_Red)_cathode

= (E°_ox)_SHE + (E°_Red)_Cu2+ _{∣ Cu}

= 0 + 0.34 V

E°_cell = 0.34V

4. Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at 25 °C.

Zn(s) ∣ Zn²⁺(aq, 1M) ∣ ∣ Cu²⁺(aq, 1M) ∣ Cu(s)

E°_Zn2+ _{∣ Zn} = −0.76 V

E°_cu2+ _{∣ Cu} = 0.34 V

E°_cell = E°_R − E°_L

= E°_Cu2+ _{∣ Cu} − E°_Zn2+ _{∣ Zn}

= 0.34 − (−0.76) = 1.10 V

 E°_cell = 1.10V

5. Write the overall redox reaction which takes place in the galvanic cell,

Pt(s) ∣ Fe²⁺(aq), Fe³⁺(aq) ∣ ∣ MnO^-₄ (aq) H⁺ (aq), Mn²⁺(aq) ∣ Pt(s)

Oxidation occur at anode

Fe²⁺ → Fe³⁺ + e⁻

Reduction occur at cathode

MnO⁻₄ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Overall reaction

5Fe²⁺ + MnO⁻ ₄ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

6. The electrochemical cell reaction of the Daniel cell is

Zn(s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s)

What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?

For Daniel cell

E_cell = E°_cell – (0.0591/2) log ( [Zn²⁺] / [Cu²⁺] )

For 1 M

E = E° − 0.0295 log [lM/1M]

E = E° − 0.0295 × 0 = E°

For 10 M

E = E° − 0.0295 log [10M/1M]

E = E° − 0.0295 log [10]

E = E° − 0.0295 × 1

E = E° − 0.0295

On increasing the ion concentration in anode compartment by 10 factor then the cell voltage decreased by the value of 0.0295.

7. A solution of a salt of metal was electrolysed for 15 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783 g. Calculate the equivalent mass of the metal.

t = 15 min = 15 × 60 = 900 sec

I = 0.15 A

 m = 0.783 g

Z = ?

m = ZIt

Z = m / It = [ 0.783g ] / [ 0.15 A × 15 × 60 sec ]

Z = 0.0058