NCERT Questions for Class 12 Chemistry Chapter 2 – Electrochemistry

Important Questions with Solutions of Class 12 Chemistry Chapter 2 – Electrochemistry

1) Electrolysis of KBr(aq) results in Br₂ at anode but won’t offer F₂. Explain why?

Ans – Oxidation happens at the anode. Since the oxidation potential is extreme, the oxidizing process is much easier. Hence, the oxidation potential of Br^–, H₂O & F^– are depicted as Br^– > H₂O > F^–.

In an aqueous KBr solution, Br^– ions are oxidized to Br₂ related to H₂O. Meanwhile, in an aqueous solution of KF, H₂O is oxidized related to F^–. So, the oxidation of H₂O at the anode provides O₂ & no F₂ is generated.

2) Explain reduction & oxidation potentials. 

Ans – Oxidation potential is the chemical species’ tendency to grab electrons loose or make them oxidized at an electrode. Meanwhile, reduction potential refers to chemical species’ property to get electrons to gain or turn them reduced at an electrode.

3) Explain why Li is the best-reducing agent & Fluorine is deemed the best oxidizing agent but not E_cell.

Ans – As per the electrochemical series, Li-ion has the least reduction potential & can act as the best-reducing agent. Meanwhile, F has a massive reduction potential & fluorine oxidizes different substances instantly. Hence, F can be the best oxidizing agent.

4) Name any 3 contrasting features between potential difference & E.M.F!

Ans –

E.M.F	POTENTIAL DIFFERENCE
1. It’s the difference between the electrode potential of 2 electrodes when no current flows inside the circuit.	1. It’s the difference of potential between the electrodes in a closed circuit.
2. Max voltage obtained from a cell.	2. Less than the highest voltage drawn out of a cell.
3. Responsible for the steady current flow.	3. Not responsible for the constant current flow.

5) After years, an electrochemical cell stops working. Define the reason!

Ans – The electrode’s reduction potential is based on the solution concentration which remains in contact. Once the cell starts working, the reactant concentration tends to decrease & will move the equilibrium backward according to Le Chatelier’s theory. Meanwhile, the equilibrium shift occurs forward, when the concentration is high on the reactant side.

When the cell works concentration in an anodic region in the cathodic part lowers & hence E°_cathode will also reduce. Now, the EMF of the cell is depicted as, E°_cell = E°_cathode – E°_anode.

A reduction in E°_cathode & a substantial increase in E°_anode refers to the cell EMF that decreases & becomes zero. This means the cell will stop functioning after a long time.

6) Define normal hydrogen electrode & its importance.

Ans – A normal hydrogen electrode is called a ‘Standard Hydrogen Electrode’ and can be deployed in half-cell potential reactions. Normal hydrogen electrodes can help to check the cell potentials using multiple electrodes & it will be a standard range of electrode potential for the thermodynamic calculation of redox potential.

7) A cell has 2 hydrogen electrodes where the negative electrode comes in contact with a solution containing 10^-6 H⁺ ions. The cell EMF is denoted as 0.118 V at 298 K. Find the concentration of the H⁺ ions during the positive electrode.

Ans – The cell reaction is referred to as,

Pt | H⁺ (10^-5 M)1atm | H⁺ (aM) | 1atm | Pt

In anode, H₂ → 2H⁺ + 2e^–

Negative polarity, [H⁺] = 10^-5 M

In cathode, 2H⁺ + 2e^– → H₂

The H⁺ ions concentration remains at the positive electrode = 10^-3 M.

8) Discuss the term electrolysis & brief the electrolysis process of:

i. Molten NaCl

Ans – Electrolysis is a chemical decomposition in aqueous solutions or a molten state via chemical change when passing an electric current. The electrolysis of molten NaCl occurs if sodium chloride is melted over 801°C, then 2 electrodes are injected into the melt & an electric current is transferred via the molten salt.

The below chemical reactions occur on the electrodes.

At Cathode: Na⁺ + e^– → Na

Then, the overall reaction will be 2NaCl → 2Na(s) + Cl₂(g)

ii. Aqueous Sodium Chloride Solution

Ans – Sodium Chloride (NaCl) dissolves with water to create Na (Sodium) & (H₂) hydrogen at their relevant electrodes.

At Cathode: H₂O(l) +2e^– → H₂(g) + 2OH^–

NaCl(aq) + H₂O(I) → Na⁺(aq) + OH^–(aq) + H₂(g) + 1/2Cl₂(g)

iii. Molten Lead Bromide

Ans – PbBr₂ (Molten lead bromide) is a proper electrolyte.

At Cathode: Pb⁺² +2e^– → Pb(s)

At Anode: 2Br^– → Br₂ +2e^–

iv. Water

Ans – The water decomposed into oxygen & hydrogen gas when passing an electric current is denoted as water electrolysis.

At Cathode: 2H₂O(l) + 2e^– → H₂(g) + 2OH^– & E° = – 0.42V

At Anode: 2H₂O → O₂(g) + 4H⁺ + 4e^– & E° = +0.82V

The overall reaction of electrolysis of water is as follows,

2H₂O(l) -> O₂ + 2H₂ & E° = – 1.24V

9) Define ‘electrolytic conduction.’ Mention the factors of electrolyte conduction & the temp effect during electrolytic conduction.

Ans – When the electric current is given to the electrolytic solutions, the ability of the solutions to enable the current is called electrolytic conductance.

Here are the factors that affect electrolytic conductance,

a. Ion concentration

b. Nature of electrolyte

c. Temp variance When there is a temperature change, the electrolyte will mix in solution. So, the increase in temp will maximize the electrolyte’s solubility & increase the electrolytic conduction.

10) Iron never rusts, even if the Zinc coating breaks in the case of a galvanized iron pipe but rusting occurs much faster if the tin coating over iron breaks down. Why?

Ans – Fe is less electropositive compared to Zn. Hence, till Zn is present on the surface of the Fe pipe, then Zn works as an anode & Fe pipe will be a cathode. Consequently, Zn restricts the rusting of Fe.

Meanwhile, Zn is extremely electropositive compared to Sn, which prevents iron from coming into direct contact with air until the Sn coating is intact. Tin cannot protect a single pore or break over since the exposed iron starts to get rusted.