Electrochemistry MCQ Quiz in मल्याळम - Objective Question with Answer for Electrochemistry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Faraday's law of electrolysis

1.) First Law of Electrolysis

It is one of the primary laws of electrolysis. It states, during electrolysis, the amount of chemical reaction which occurs at any electrode under the influence of electrical energy is proportional to the quantity of electricity passed through the electrolyte.

2.) Second Law of Electrolysis

Faraday’s second law of electrolysis states that if the same amount of electricity is passed through different electrolytes, the masses of ions deposited at the electrodes are directly proportional to their chemical equivalents.

Newton's law of motion

1.) Newton’s first law: the law of inertia

It states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. 

2.) Newton’s second law: F = ma

It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it. 

3.) Newton’s third law: the law of action and reaction

It states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction. 

Faraday's law of induction

Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced. Likewise, if the conductor circuit is closed, a current is induced, which is called induced current.

Gauss's law

It states that the electric flux (Φ) across any closed surface is proportional to the net electric charge enclosed by the surface.

Explanation:- 

• Due to the low reactivity of copper, it does not liberate hydrogen gas when it reacts with acid. It forms green corrosion with air because it easily reacts with moisture present in the air.
• Metal having less reactivity than hydrogen do not react with dilute acid. 
• Order of Reactivity:- Mg < Fe < Cu <  Ag.
• Reactivity Series of Metal:- It is commonly known as activity series is an arrangement of in decreasing order of their reactivity.

Additional Information 

The correct answer Molar conductivity increases with a decrease in concentration.

Molar conductivity:

It is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length. 

Molar conductivity has the SI unit S m² mol⁻¹.

Molar conductivity increases with a decrease in concentration.

 ⇒ Λ_m × M = k × 1000

Where,

Λm - Molar conductivity

V - Volume , k - Specific conductance

M is the molarity

→ As seen in the above equation we can arrive at a conclusion that Λ _m⁰ is inversely proportional to Molarity. 
→ And hence if we increase concentration of solution molar conductivity decreases and vice-versa.

 Additional Information

Specific conductance:

The conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of crosssection 1 sq. cm. Specific conductance is the inverse of resistivity.

⇒ k = 1/p 

where, p - Resistivity

explanation:

Given:

 Conductivity, k = 0.146 x 10-3 S cm-1

Resistance, R = 1000

Cell constant = k x R

 = 0.146 X 0-3 X 1000

 = 0.146 cm^-1

 Important Points

EMF of cell:

\(\rm E^0_{cell}=\begin{bmatrix}\rm Standard \ reduction\\\ \rm Potential \ of\ cathode\end{bmatrix}-\begin{bmatrix}\rm Standard \ oxidation\\\ \rm Potential \ of\ anode\end{bmatrix}\)

\(\rm E^0_{cell}=E^0_{cathode}-E^0_{anode}=E^0_{right}-E^0_{left}\)

• Nernst Equation:

​\(\rm M^{n+}_{(aq)}+ne^-\longrightarrow M_{(s)}\)

\(\rm E=E^0-\frac{RT}{nF}ln \frac{[M_{(s)}]}{[M^{n+}_{(aq)}]}\)

\(\rm E=E^0-\frac{2.303RT}{nF}log \frac{[M_{(s)}]}{[M^{n+}_{(aq)}]}\)

• Gibbs Free Energy Change:

​-ΔG = nFE

• Faraday First Law of Electrolysis:

​
m ∝ It or m = ZIt

Where

I = Current in amperes

t = time in seconds

m = mass of the primary product in grams

Z = constant of proportionality

(Electrochemical equivalent). It is the mass of a substance liberated by 1 ampere second of a current (1 coulomb).

The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The conductivity of electrolytic (ionic) solutions depends on :

(i) the nature of the electrolyte added

(ii) size of the ions produced and their solvation

(iii) the nature of the solvent and its viscosity

(iv) concentration of the electrolyte

(v) temperature (it increases with the increase of temperature)

CONCEPT:

Standard EMF of a Galvanic Cell

• The standard EMF (electromagnetic force) of a galvanic cell is calculated by subtracting the standard electrode potential of the anode (oxidation half-reaction) from that of the cathode (reduction half-reaction).
• The formula to calculate the EMF is:
  • EMF = E°_cathode - E°_anode

CALCULATION:

• For a galvanic cell with zinc and copper electrodes:
  • The cathode is copper (reduction occurs): E°_cathode = +0.34 V
  • The anode is zinc (oxidation occurs): E°_anode = -0.76 V
• Applying the formula:
  • EMF = 0.34 V - (-0.76 V) = 0.34 V + 0.76 V = 1.10 V

CONCLUSION:

• Correct option: Option 1 - 1.10 V

For redox change

\(\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}\)

n-factor = 5

n_eq = nmol × n-factor

= 1 × 5

= 5

Charge required in Faraday = Number of n_eq

= 5 F

Explanation:-

Molar Conductance:-

• Molar conductance is the conductance property of a solution containing one mole of electrolyte or a function of a solution's ionic strength or salt concentration. As a result, it is not a constant.

We know,

⇒∧° m(BaCl2) →  ∧° m(Ba²⁺)  +  2 ∧° m(Cl^-)   --- [1]

⇒∧° m (NaOH) →  ∧° m (Na⁺) +  ∧° m (OH^-)   ---- [2]

⇒∧° m (NaCl)  →  ∧° m (Na⁺)  +∧° m (Cl^-)   ---- [3]

For the reaction -  Ba²⁺  + 2 OH^-  → Ba(OH)₂ , 

⇒ Eq [1] +  2 × Eq [3] - 2 × Eq [2]

⇒ ∧° m(BaCl2) →  ∧° m(Ba2+)  +  2 ∧° m(Cl-)    + 2 x ∧° m (NaCl)  →  ∧° m (Na+)  +∧° m (Cl-)   - 2 x ∧° m (NaOH) →  ∧° m (Na+) +  ∧° m (OH-) 

⇒ (∧° m) for Ba(OH)₂ =  280.0 × 10-4 S m2mol^-1 + 2 x 248.1 × 10-4 S m2mol-1  - 2 x  126.5 × 10-4 S m2mol-1.

∴ (∧° m) for Ba(OH)2 = 523.2 × 10^-4 S m2mol-1.

Explanation-

• The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts.
• The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell.
• For a general electrochemical reaction of the type: aA + bB → cC + dD

Nernst equation can be written as:

\({E_{({M^{n + }}/M)}} = {E^ - }_{({M^{n + }}/M)} - \frac{{RT}}{{nF}}\ln \frac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}\)

• Where, R is the gas constant (8.314 JK–1 mol–1), F is the Faraday constant (96487 C mol–1), and T is the temperature in kelvin.
• By putting the values of constant the above equation will be
• \({E_{({M^{n + }}/M)}} = {E^ - }_{({M^{n + }}/M)} - \frac{{0.0591}}{n}\log \frac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}} \)
• E^o_cell = (E^o_red)_c - (E^o_red)_a

Given data and Calculation:

Cd + Fe²⁺ → Cd²⁺ + Fe

So number of electrons participating = n = 2

E0{Fe2 +/Fe} = - 0.44V; E0Cd2 +/Cd = - 0.40V

E^o_cell = -0.40 - (-0.44) = -0.04

\({E_{cell}} = 0.04 - \frac{{0.0591}}{2}\log \frac{0.1}{0.001} \\ {E_{cell}} = 0.04 - \frac{{0.0591}}{2}\log{10^{2}} \) 

=- 0.0191 V

CONCEPT:

Gibbs Free Energy (ΔG^o) and Cell Potential (E^o_cell)

• The relationship between the Gibbs free energy change (ΔG^o) and the standard cell potential (E^o_cell) in an electrochemical cell is given by the equation:

  ΔG^o = -nFE^o_cell

• Where:
  • ΔG^o: Standard Gibbs free energy change
  • n: Number of moles of electrons transferred in the cell reaction
  • F: Faraday's constant (96485 C/mol)
  • E^o_cell: Standard cell potential

EXPLANATION:

• In the given cell:

  Sn(s)|Sn²⁺(1M)||Ag⁺(1M)|Ag(s)

  • Tin (Sn) is oxidized to Sn²⁺, so Sn is the anode.
  • Silver (Ag⁺) is reduced to Ag, so Ag is the cathode.
• Given data:
  • E^o_cell = 0.90 V
  • The reaction involves the transfer of 2 moles of electrons.
• Using the equation for Gibbs free energy:
  • ΔG^o = -nFE^o_cell
  • n = 2, F = 96485 C/mol, and E^o_cell = 0.90 V
  • ΔG^o = -2 × 96485 × 0.90
  • ΔG^o = -173717 J/mol or -173.7 kJ/mol

Therefore, the Gibbs free energy change (ΔG^o) for the cell is -173.7 kJ/mol.