To determine which electrolyte can be used to obtain \( \text{H}_2\text{S}_2\text{O}_8 \) (peroxydisulfuric acid) by the process of electrolysis, we need to understand the conditions required for its formation:
1. Principle of Electrolysis: Electrolysis involves the decomposition of compounds using an electric current. The electrolyte's nature significantly affects the products formed.
2. Peroxydisulfuric Acid Formation: This compound is specifically formed when concentrated sulfuric acid is electrolyzed. The equation for the formation of \( \text{H}_2\text{S}_2\text{O}_8 \) is as follows:
2 \( \text{HSO}_4^- \) (aq) → \( \text{H}_2\text{S}_2\text{O}_8 \) (aq) + 2 \( e^- \)
3. Options Analysis:
| Option | Suitability |
| Dilute solution of sodium sulphate | Not suitable, lacks sufficient sulfate ions and concentration. |
| Dilute solution of sulphuric acid | Not suitable due to low concentration. |
| Concentrated solution of sulphuric acid | Suitable; provides high concentration of sulfate ions for forming \( \text{H}_2\text{S}_2\text{O}_8 \). |
| Acidified dilute solution of sodium sulphate | Not suitable, lacks adequate concentration of \( \text{HSO}_4^- \) ions. |
The process's efficiency in forming \( \text{H}_2\text{S}_2\text{O}_8 \) increases with the concentration of \( \text{HSO}_4^- \) ions provided by concentrated sulfuric acid, which makes it the correct choice.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)