The value of the constant \(A\) in the Debye - Huckel -Onsager equation, \(\lambda_{ m }=\) \(\lambda_{m}^{0}-A \sqrt{C}\) depends on temperature, charges of the ions, the dielectric constant of the solvent and also the viscosity of the solvent. Here, since we are not given the solvent, we will assume that the solvent is the same for each case. Here, the deciding factor is the charge on the ions. Again, in \(NH _{4} Cl\) and \(NaBr\), the charges on the ions are the same.
Hence the pair of electrolytes that will possess the same value for the constant (A) will be \(NH _{4} Cl , NaBr\).
The Debye-Hückel-Onsager equation relates the molar conductivity (\( \lambda_m \)) of an electrolyte solution to its concentration (\( C \)): \[ \lambda_m = \lambda^\circ_m - A \sqrt{C} \] where:
\( \lambda_m \) is the molar conductivity at concentration \( C \),
\( \lambda^\circ_m \) is the limiting molar conductivity (when concentration approaches zero),
\( A \) is a constant that depends on the nature of the electrolyte and the solvent. The constant \( A \) is influenced by the ionic strength of the solution and the type of ions present.
For a given type of solvent (water, in this case), the value of \( A \) is the same for pairs of electrolytes that have the same type of ion (monovalent or divalent). The key observation is that the electrolytes that are pairwise similar in their ion characteristics tend to have similar values of \( A \).
Looking at the options:
MgSO₄ and Na₂SO₄ both have divalent cations (Mg²⁺ and Na⁺), and therefore, the constant \( A \) will be the same for both, as their ionic interactions are similar.
Thus, the correct answer is (B).
If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).
Consider the following
Statement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product.
Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
Electrolysis is the process by which an element is decomposed and undergoes some chemical change under the influence of any electric current. The first-ever electrolysis was executed out by Sir Humphrey Davey in the year 1808. Electrolysis can occur in both Galvanic cells and Electrolytic cells.
Read More: Products of Electrolysis