Unit 10: Electrochemistry — Short Questions

Ans. Electrical double layer is formed at the interface between a metal electrode and the electrolyte solution due to the separation of charges. When a metal is dipped into its salt solution, metal atoms lose electrons and become positive ions, which move into the solution, while the electrons remain on the electrode. This creates a layer of positive ions in the solution and negative charge on the metal surface, forming two layers of opposite charges hence called the electrical double layer.
Reason: It is formed to maintain electrochemical equilibrium between the electrode and the electrolyte.

Ans. The electrode potential of copper (Cu) is called reduction potential because it is measured for the reaction in which Cu²⁺ ions gain electrons to form Cu metal:
Cu²₍ₐq₎ + 2e⁻ ⟶ Cu₍ₛ₎
This process involves the gain of electrons, which is defined as reduction, so the potential is called reduction potential.

Ans. The advantages of a salt bridge in a galvanic cell are:
(i) Maintains electrical neutrality by allowing the flow of ions between the two half-cells.
(ii) Completes the electrical circuit to allow continuous flow of electrons.
(iii) Prevents mixing of different solutions, which could cause direct chemical reaction and stop the cell from working.

Ans. The feasibility of a chemical reaction can be predicted using the cell voltage (E°ₑₗₗ).
If the E°ₑₗₗ is positive, the reaction is spontaneous and feasible.
If the E°ₑₗₗ is negative, the reaction is non-spontaneous and not feasible.
E°ₑₗₗ = E°cₐₜₕₒdₑ − E°ₐₙₒdₑ

Ans. During electrolysis of aqueous NaCl, Na⁺ ions are not liberated at the cathode because water is reduced more easily than sodium ions.
Water has a higher reduction potential than Na⁺, so H₂ gas is produced instead of sodium metal.
2H₂O+2e⁻ ⟶ H₂₍ₘ₎+2OH⁻

Ans. (i) CrCl₃
Let the oxidation number of Cr = x
Chlorine (Cl) has an oxidation number of -1
x + 3(−1)=0⟹x=+3
Oxidation number of Cr = +3

(ii) Cr₂(SO₄)₃
Let oxidation number of Cr = x
Sulfate ion (SO₄²⁻) has a charge of -2
Total charge from 3 sulfate ions = 3 × (-2) = -6
So for the compound Cr₂(SO₄)₃:
2x+(−6)=0⟹2x=+6⟹x=+3
Oxidation number of Cr = +3

(iii) Cr₂O₃²⁻ (Dichromate ion)
Let oxidation number of Cr = x
Oxygen (O) has an oxidation number of -2
Total charge from 7 oxygen atoms
= 7 × (-2) = -14
Total charge on the ion = -2
2x + (−14) = −2 ⟹ 2x = +12 ⟹ x = +6
Oxidation number of Cr = +6

Ans. (i) Copper is added to a solution of magnesium sulfate (MgSO₄):
Reaction prediction:
Since copper (Cu) is less reactive than magnesium (Mg), no reaction will occur.
No reaction, because copper cannot displace magnesium from its salt solution.

(ii) Iron is added to a dilute solution of hydrochloric acid (HCl):
Reaction prediction:
Iron (Fe) is more reactive than hydrogen, so it will react with HCl to produce hydrogen gas.
Fe + 2HCl ⟶ FeCl₂ + H₂↑

Ans. Metals higher in the activity series are more reactive and can easily lose electrons (oxidize). When these metals react with acids, they displace hydrogen ions (H⁺) from the acid because they have a stronger tendency to oxidize compared to hydrogen.
On the other hand, metals lower in the activity series have a weaker tendency to lose electrons and cannot displace hydrogen ions from the acid. These metals are less reactive, and thus hydrogen is not displaced in reactions with acids.
In simple terms:

• Higher activity series metals (e.g., Mg, Zn): More reactive → can displace H⁺ ions.
• Lower activity series metals (e.g., Cu, Pt): Less reactive → cannot displace H⁺ ions.

Ans. To calculate the number of Faradays required to deposit the given masses of Ag⁺, Cu²⁺, and Al³⁺, we use Faraday's Laws of Electrolysis. The formula to calculate the number of Faradays is:
Faradays = Mass of substance / (Molar mass of substance × n)

Where

• n is the number of electrons involved in the reduction process for each ion.
• Molar mass is the molar mass of the substance.

(i) For Ag⁺ (Silver ion)

• Molar mass of Ag = 108 g/mol
• Ag⁺ gains 1 electron to form Ag (s), so n = 1 Faradays for Ag = 108g / (108 g/mol × 1) = 1 Faradays

(ii) For Cu²⁺ (Copper ion)

• Molar mass of Cu = 63.5 g/mol
• Cu²⁺ gains 2 electrons to form Cu (s), so n = 2 Faradays for Cu = 63.5g / (63.5 g / mol × 2) = 127 / 127 = 0.5 Faraday

(iii) For Al³⁺ (Aluminum ion)

• Molar mass of Al = 27 g/mol
• Al³⁺ gains 3 electrons to form Al (s), so n = 3 Faradays for Al = 27g / (27 g / mol × 3) = 0.333 Faradays

The total number of Faradays required to deposit 108 g of Ag⁺, 63.5 g of Cu²⁺, and 27 g of Al³⁺ is 1.833 Faradays.

Ans. To calculate Avogadro's number (Nₐ) from electrolysis data, we use Faraday's Laws of Electrolysis and relate the total charge passed to the number of atoms deposited.

Given Data

• Current (I) = 0.500 A
• Time (t) = 30.0 minutes = 30 × 60 = 1800 seconds
• Mass of silver deposited (m) = 0.503 g
• Molar mass of Ag (M) = 107.87 g/mol

Step 1 Calculate total charge (Q) passed
Q = I × t = 0.500 A × 1800 s = 900 C

Step 2 Calculate moles of Ag deposited:
Moles of Ag = Mass / Molar Mass
Moles of Ag = 0.503 / 107.87
Moles of Ag = 4.664×10⁻³ mol

Step 3 Calculate moles of electrons used
Since 1 mole of Ag⁺ requires 1 mole of electrons:
Moles of electrons = Moles of Ag
= 4.664 × 10⁻³ mol

Step 4 Use charge to find charge per mole (Faraday constant)
Charge per moles (F) = Q / Moles of Ag
= 900 / 4.664×10⁻³
= 1.93 × 10⁵ C/mol

Step 5 Use Faraday's constant to find Avogadro's number
Faraday's constant (F) = Nₐ × e
e = 1.602 × 10⁻¹⁹ C (charge of one electron)
Nₐ = F / e = 1.93×10⁵ / 1.602×10⁻¹⁹ = 6.02×10²³

Ans. Given Standard Electrode Potentials:

• Ni²⁽ₐq₎ + 2e ⇌ Ni₍ₛ₎ E° = −0.25 V
• Co²⁽ₐq₎ + 2e ⇌ Co₍ₛ₎ E° = −0.28 V

(i) Identify anode and cathode:

• In a galvanic (voltaic) cell, oxidation occurs at the anode and reduction at the cathode.
• The electrode with lower (more negative) reduction potential is more likely to oxidize (lose electrons).
• Here, Co has a lower E° (−0.28 V) than Ni (−0.25 V), so Co is oxidized and acts as the anode.

Answer

• Anode: Cobalt (Co)
• Cathode: Nickel (Ni)

Justification Co has a lower (more negative) standard reduction potential, so it is more easily oxidized than Ni.

(ii) Write the balanced overall cell reaction:

• Anode (oxidation): Co₍ₛ₎ ⟶ Co²⁽ₐq₎ + 2e

• Cathode (reduction): Ni²⁽ₐq₎ + 2e ⟶ Ni₍ₛ₎

Overall cell reaction Co₍ₛ₎ + Ni²⁽ₐq₎ ⟶ Co²⁽ₐq₎ + Ni₍ₛ₎

(iii) Calculate the standard cell potential (E°ₑₗₗ):
E°ₑₗₗ = E°cₐₜₕₒdₑ − E°ₐₙₒdₑ
E°ₑₗₗ = (− 0.25 V) − (− 0.28 V) = +0.53 V

Answer E°ₑₗₗ = +0.53 V

Ans. Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It includes the study of redox reactions where electrons are transferred.
Example: Electrolysis of water produces hydrogen and oxygen gases using electricity.

Ans. A redox reaction is a chemical reaction that involves the transfer of electrons. One substance is oxidized (loses electrons), and another is reduced (gains electrons).
Example:
Zn₍ₛ₎ + Cu²⁽ₐq₎ ⟶ Zn²⁽ₐq₎ + Cu₍ₛ₎
(Zn is oxidized; Cu²⁺ is reduced)

Ans. Oxidation is the loss of electrons by an atom or ion.
Example:
Na₍ₛ₎ ⟶ Na⁺₍ₐq₎ + e⁻
(Sodium loses one electron)

Ans. Reduction is the gain of electrons by an atom or ion.
Example:
Cl₂₍ₐq₎ + 2e⁻ ⟶ 2Cl⁻₍ₐq₎
(Chlorine gains electrons)

Ans. Photosynthesis is a redox reaction which provides food for the entire planet, and another one is respiration that keeps us alive, both are redox reactions.

Ans. A device that converts chemical energy into electrical energy through redox reactions.
Example: Daniell cell (Zn-Cu cell).

Ans. A salt bridge maintains electrical neutrality by allowing the movement of ions between the two half-cells.

Ans. Conductors (usually metals) where oxidation or reduction occurs.

• Anode: where oxidation occurs
• Cathode: where reduction occurs

Ans. A cell that generates electrical energy from a spontaneous redox reaction.
Example: Daniell cell:
Zn₍ₛ₎ / Zn²⁺₍ₐq₎ || Cu²⁺₍ₐq₎ / Cu₍ₛ₎

Ans. The anode is negative because it loses electrons (oxidation).

Ans. Electrons flow from anode to cathode. From where it can be used in external circuit.

Ans. It is a classic example of a galvanic cell that demonstrates conversion of chemical to electrical energy.

Ans. A device where electrical energy is used to drive a non-spontaneous chemical reaction.
Example: Electrolysis of molten NaCl.

Ans. Anode is positive because it attracts negative ions (oxidation still occurs here).

Ans.

• At cathode: Na⁺₍ₗ₎ + e⁻ ⟶ Na₍ₛ₎
• At anode: Cl₍ₗ₎ ⟶ ½ Cl₂₍ₘ₎ + 2e⁻

Ans. The process of depositing a metal layer on an object using an electrolytic cell.
Example: Electroplating silver on a spoon.

Ans. The potential difference between a metal electrode and its ion solution.

Ans. The electrode potential measured under standard conditions: 1M, 1 atm, 25°C.

Ans. A reference electrode with E° = 0.00 V. used to measure other electrode potentials.

Ans. Because it is stable, reversible, and easy to reproduce.

Ans. The electromotive force is the potential difference between the electrodes of a cell when no current is drawn.
E°ₑₗₗ = E°cₐₜₕₒdₑ − E°ₐₙₒdₑ

Ans. Positive EMF → spontaneous reaction

• Negative EMF → non-spontaneous reaction

Ans.
Anode/Anode sol. || Cathode sol. / Cathode
Example:
Zn₍ₛ₎ / Zn²⁺₍ₐq₎ || Cu²⁺₍ₐq₎ / Cu₍ₛ₎

Ans. A list of elements arranged in order of increasing standard electrode potentials.

Ans. A metal higher in the series can displace a lower metal from its salt solution.
Example:
Zn₍ₛ₎ + CuSO₄ ⟶ ZnSO₄ + Cu₍ₛ₎

Ans. Because it has a very negative E° value (−2.92 V), indicating high tendency to lose electrons.

Ans. In batteries such as dry cells, lead-acid batteries, and lithium-ion cells.

Ans. A galvanic cell using MnO₂ as cathode and Zn as anode with an electrolyte paste.

Ans. A redox process where metals (like iron) get oxidized due to environmental exposure.
Example:
Fe₍ₛ₎ ⟶ Fe³⁺₍ₐq₎ + 2e⁻

Ans.

• Conductor: Material that allows free flow of electrons (e.g., metals)
• Electrolyte: Compound that conducts electricity through ion movement in solution or molten state (e.g., NaCl solution).

Ans. The mass of substance deposited or liberated is directly proportional to the quantity of electricity passed.

Ans.

• Hydrogen shows -1 oxidation state with all metals.
• Hydrogen shows +1 oxidation state with all non-metals.
• Oxygen shows negative oxidation state with all metals.
• Oxygen shows negative oxidation state with all non-metals except fluorine.
• Oxygen shows -2, -1 and -1/2 in H₂O, Na₂O₂, and KO₂, respectively.

Ans.
(a) If an element in a species undergoes only increase in oxidation number or decrease in oxidation number in a reaction. The specie containing such element is written once on the RHS of the equation.
Zn₍ₛ₎ + HCl₍ₐq₎ ⟶ ZnCl₂₍ₐq₎ + H₂₍ₘ₎

(b) If an element in a species undergoes both increase in oxidation number or decrease in oxidation number simultaneously in a reaction (as in case of self-redox or disproportionation reaction). The specie containing such element is written twice on the RHS of the equation. First for increase in oxidation number while second is written for decrease in oxidation number.
Cl₂ + NaOH ⟶ NaCl + H₂O + NaOCl

(c) If an element in a species undergoes increase or decrease in oxidation number as well as no change in oxidation number in a reaction. The specie containing such element is also written twice. First, for change in oxidation number while second is for no change in oxidation number.
HNO₃ + Cu + HNO₃ ⟶ Cu(NO₃)₂ + H₂O + NaOCl
HCl + K₂Cr₂O₇ + HCl ⟶ Cl₂ + KCl + CrCl + H₂O

Ans. Sequence of the balancing the chemical equation using inspection (hit, and trial) method.
Firstly, balance atoms of all elements except nitrogen, oxygen and hydrogen. Secondly, balance atoms of nitrogen (if in the equation). Thirdly, balance atoms of oxygen (if in the equation). Lastly, balance atoms of hydrogen (if in the equation).

Ans. See Q15. from theory.

Ans. See Q8. from theory.

Ans. See Q12. from theory.

Ans. See Q10. from theory.

Solution Given:

• Standard electrode potential, E° = − 0.76 V
• Concentration of Zn²⁺ = 0.010 mol dm⁻³
• Temperature, T = 298 K
• Gas constant, R = 8.31 JK⁻¹ mol⁻¹
• Faraday constant, F = 96500 C mol⁻¹

Step 1 Write the half-cell reaction
Zn²⁺₍ₐq₎ + 2e ⟶ Zn₍ₛ₎
Number of electrons transferred, n = 2.

Step 2 Write the Nernst equation
E = E° − (RT/nF) ln Q

Here, Q is the reaction quotient for the half reaction:
Q = 1/[Zn²⁺] 0.010 = 100
(Since solid zinc activity = 1)

Step 3 Plug values into the Nerst equation
E = −0.76 − (8.31)(298)/(2)(96500) ln 100

Calculate the constant term
(8.31)(298)/(2)(96500) = 2477.66/193000 = 0.012837V

Calculate ln 100
ln 100 = ln(10²) = 2 ln 10 = 2 × 2.303 = 4.606

Step 4 Calculate the potential
E = −0.76 − (0.012837 × 4.606)

Answer E = −0.82V

Solution Given:

• Current, I = 2.00 A
• Time, t = 30.0 minutes = 30 × 60 = 1800 seconds
• Molar mass of Cu = 63.5 g/mol
• Faraday constant, F = 96500 C/mol

Step 1 Write the cathode half-reaction
Cu²⁺+2e⁻ → Cu₍ₛ₎
This means 2 moles of electrons are required to deposit 1 mole of Cu.

Step 2 Calculate total charge passed
Q = I × t = 2.00 × 1800 = 3600 C

Step 3 Calculate moles of electrons transferred
Moles of electrons = Q/F = 3600/96500 ≈ 0.0373 mole

Step 4 Use mole ratio to find moles of copper deposited
From the reaction:
2 mol e⁻ → 1 mol Cu

So, Moles of Cu deposited = 0.0373/2 = 0.01865 mole

Step 5 Calculate mass of copper deposited
Mass = moles × molar mass = 0.01865 × 63.5 = 1.18 g

Answer 1.18 g of copper is deposited at the cathode.

Ans. (a) Write the balanced overall cell reaction.
First, identify the half-reactions
(1) Standard Hydrogen Electrode (SHE)
2H⁺₍ₐq₎ + 2e ⟶ H₂₍ₘ₎ (E° = 0.00 V)

(2) Nickel Half-cell (reverse of reduction):
Since nickel is the negative terminal, it acts as the anode (oxidation occurs), so the reaction is:
Ni₍ₛ₎ ⟶ Ni²⁺₍ₐq₎ + 2e

Overall Cell Reaction Add the two half-reactions:
Ni₍ₛ₎ + 2H⁺₍ₐq₎ ⟶ Ni²⁺₍ₐq₎ + H₂₍ₘ₎

Answer for (a) Ni₍ₛ₎ + 2H⁺₍ₐq₎ ⟶ Ni²⁺₍ₐq₎ + H₂₍ₛ₎

(b) Determine the standard electrode potential (E°) of the Ni²⁺/Ni₍ₛ₎ half-cell.
Given:

• Cell potential, Eₑₗₗ = 0.25 V
• SHE is always assigned E° = 0.00 V
• Nickel is the anode, so: E°ₑₗₗ = E°cₐₜₕₒdₑ − E°ₐₙₒdₑ

Substitute values 0.25 = 0.00 − E°ₙᵢ²⁺/ₙᵢ
Solve for 0.25 = 0.00 − E°ₙᵢ²⁺/ₙᵢ
E°ₙᵢ²⁺/ₙᵢ = −0.25V

Answer for (b) E°ₙᵢ²⁺/ₙᵢ = −0.25V

(c) Identify which electrode is the anode and which is the cathode.
From the problem:

• Nickel electrode is negative terminal ⇒ anode
• SHE is positive terminal ⇒ cathode

Answer for (c) Anode: Ni₍ₛ₎ Cathode: SHE(H₂ / H⁺)