Electrochemistry & Redox: 35 MCQs with Explanations

In S₂O₃²⁻, one S has oxidation state 0 (central S) and the other has +6 (like sulfate S), giving average oxidation state of +2 per S atom. Structure: [O₃S-S]²⁻ where terminal S is +5? Actually standard knowledge: average oxidation state of S in thiosulfate is +2.

By convention, the standard hydrogen electrode (SHE) is assigned a potential of 0.00 V at all temperatures.

Lithium has the most negative electrode potential (E° = -3.04 V), making it the strongest reducing agent despite having highest ionization energy (due to high hydration energy of Li⁺).

Let oxidation state of Mn = x. For KMnO₄: +1 + x + 4(-2) = 0 → x - 7 = 0 → x = +7.

Let oxidation state of Cr = x. For K₂CrO₄: 2(+1) + x + 4(-2) = 0 → x - 6 = 0 → x = +6.

Let oxidation state of N = x. For NO₃⁻: x + 3(-2) = -1 → x - 6 = -1 → x = +5.

Let oxidation state of Fe = x. For K₃[Fe(CN)₆]: 3(+1) + x + 6(-1) = 0 → x - 3 = 0 → x = +3.

Lithium has the most negative electrode potential (E° = -3.04 V) among alkali metals, making it the strongest reducing agent despite having the highest ionization energy (due to high hydration energy of Li⁺).

Let oxidation state of S be x. For H₂SO₄: 2(+1) + x + 4(-2) = 0 → 2 + x - 8 = 0 → x = +6.

Let oxidation state of Mn be x. For KMnO₄: 1(+1) + x + 4(-2) = 0 → 1 + x - 8 = 0 → x = +7.

Let oxidation state of N be x. For NH₄⁺: x + 4(+1) = +1 → x + 4 = 1 → x = -3.

Let oxidation state of Fe be x. For K₄[Fe(CN)₆]: 4(+1) + x + 6(-1) = 0 → 4 + x - 6 = 0 → x = +2. (CN⁻ has -1 charge).

E°cell = E°cathode - E°anode = E°Cu - E°Zn = 0.34 - (-0.76) = 1.10 V.

Let the oxidation state of N be x. H is +1, O is -2. So, (+1) + x + 3(-2) = 0 => 1 + x -6 = 0 => x = +5.

The catalytic activity of transition metals is often linked to their ability to adopt multiple oxidation states and provide a surface for adsorption, providing alternative reaction pathways with lower activation energy.

Primary alcohols are first oxidized to aldehydes, which are then further oxidized to carboxylic acids under strong oxidizing conditions.

The salt bridge contains an inert electrolyte. It maintains electrical neutrality in both half-cells by allowing the flow of anions towards the anode and cations towards the cathode, thus completing the circuit.

By convention, the Standard Hydrogen Electrode (SHE) is assigned an electrode potential of exactly 0.00 volts. It serves as a reference for measuring all other standard electrode potentials.

This is the statement of Faraday's First Law of Electrolysis. It relates the mass of a substance produced at an electrode to the electric current used.

Let oxidation state of Cr be x. K is +1, O is -2. For K₂Cr₂O₇: 2(+1) + 2x + 7(-2) = 0 => 2 + 2x -14 = 0 => 2x = 12 => x = +6.

In the electrolysis of aqueous NaCl, at the cathode, water is reduced more easily than Na⁺ ions. The reduction reaction is 2H₂O + 2e⁻ → H₂ + 2OH⁻. Hence, hydrogen gas is liberated at the cathode.

Acidified K₂Cr₂O₇ is an oxidizing agent. Oxidation of primary alcohols like ethanol yields aldehydes (ethanal) upon mild oxidation. Further oxidation yields carboxylic acids.

Magnesium has a +2 oxidation state. Nitrogen (nitride ion) has a -3 oxidation state. To balance the charges, we need three Mg²⁺ ions (+6) and two N³⁻ ions (-6), giving the formula Mg₃N₂.

Electrolysis of acidified water: At anode (oxidation): 2H₂O → O₂ + 4H⁺ + 4e⁻. Oxygen gas is liberated at the anode. Hydrogen gas is liberated at the cathode.

Aluminum is extracted from its ore (bauxite) by electrolytic reduction (Hall-Héroult process) because it is a highly reactive metal and cannot be reduced by common reducing agents like carbon.

By convention, standard reduction potential of 2H⁺ + 2e⁻ → H₂ is defined as 0.00 V at all temperatures.

m = ZQ = (E/F)Q, where Z = electrochemical equivalent, E = equivalent weight, Q = charge passed.

Aluminum is extracted by Hall-Héroult process: electrolysis of molten Al₂O₃ dissolved in cryolite (Na₃AlF₆).

Rusting involves formation of galvanic cells on iron surface where Fe oxidizes at anode and O₂ reduces at cathode.

Boron exhibits +3 oxidation state in all its compounds (e.g., B₂O₃, BF₃) due to its small size and high ionization energy.

White phosphorus undergoes slow oxidation in air producing chemiluminescence (glow).

Lead-acid battery uses ~30% H₂SO₄ as electrolyte. Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.

Ligands named alphabetically (ammine before chlorido); oxidation state of Co = +3; counter ions named last.

Let oxidation state of Fe = x. 4(+1) + x + 6(-1) = 0 ⇒ 4 + x - 6 = 0 ⇒ x = +2.

Pyridinium chlorochromate (PCC) in anhydrous conditions oxidizes 1° alcohols to aldehydes without further oxidation to carboxylic acids.