As it is given that this reaction is in acidic medium, that means we have H+ on the reactant side so to balance Oxygen we have to add water on the product side instead of adding just Oxygen.

The metals, present below hydrogen in the electrochemical series, cannot liberate hydrogen from the dilute acids.

Among the given metals only Ag is present below hydrogen in electrochemical series, so it does not evolve hydrogen with dil HCl.

Ag−I⁻dilHCl ⟶ No reaction

For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of Fe⁺² is less than that of [Fe(CN)₆​]⁴⁻. Therefore, Fe⁺² is stronger oxidizing agent than [Fe(CN)₆​]⁴⁻. Also, the stronger oxidizing agent should easily reduce itself. Here, Fe⁺³ is easily reduced than Fe⁺². Therefore, among all the four, Fe⁺³ is the stronger oxidizing agent.

Option A is correct.

log₁₀Kp = 8 - 6400/T.

Convert temperature: T = 527 °C = 527 + 273 = 800 K.

Substitute T = 800 K into the formula: log₁₀Kp = 8 - 6400/800 = 8 - 8 = 0.

Therefore Kp = 10⁰ = 1.

For an equilibrium that involves only pure solids and CO₂(g), the activities of solids are unity, so Kp = p_CO2 (with pressure in atm). Hence p_CO2 = 1 atm, matching option A.

Electrolysis of H₂SO₄

Cathode:-

2H⁺  +  2e^-  -  H₂    (Reduction)

Anode:-

4OH^-  -  2H₂O  +  O₂  + 4e^-  (Oxidation)

OH^- ions discharged at anode.

(i) Oxidation state of element in its free state is zero.

(ii) Sum of oxidation states of all atoms in compound is zero.

O.N of S in S₈=0;    O.N of S in S₂F₂=+1

O.N of S in H₂S = -2

Boron in B₂H₆ is electron deficient

Reduction potential means to accept electrons to reduce oneself.

 A + e^- → A^- ∆E_reduction = +ve value

Since, the reduction potential is negative, it means that the reaction will reverse to make ∆E value +ve. So the reaction becomes,

A → A⁺ + e^- 

This becomes oxidation of A. So oxidation of A will be easy.

The correct answer is Option B - A state where the concentrations of reactants and products remain constant with time

This describes a dynamic equilibrium in a closed system, where forward and reverse changes continue but balance each other.

At equilibrium the rates of the forward and reverse reactions are equal, and as a result the concentrations remain constant with time.

Equilibrium does not mean the reactions have stopped; microscopic conversion of reactants to products and products to reactants continues, but with no net change in concentrations.

The amounts of reactants and products at equilibrium are not necessarily equal; their relative values depend on the equilibrium constant.

Kc = [C]^c[D]^d / [A]^a[B]^b

If Kc > 1, products are favored at equilibrium; if Kc < 1, reactants are favored. This shows equal concentrations are not required for equilibrium.

Therefore, Option B is the correct description of chemical equilibrium.