CBSE 2025 · Region 1 · Set 1 · Q23 · 3 marks
(a)State the following : $\displaystyle 3$(i)Kohlrausch law of independent migration of ions and(ii)Faraday's first law of electrolysis.(b)Using $\displaystyle \mathrm{E}_{\text {values }}^{\circ}$ of X and Y given below, predict which is better for coating the surface of iron to prevent corrosion and why ? Given $\displaystyle \mathrm{E}_{\mathrm{X}^{2+} / \mathrm{X}}^{\circ}=-2.36 \mathrm{~V}$, \[\begin{aligned} & \mathrm{E}_{\mathrm{Y}^{2+} / \mathrm{Y}}^{\circ}=-0.14 \mathrm{~V}, \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V} \end{aligned} \]
(a)
State the following : $\displaystyle 3$
(i)
Kohlrausch law of independent migration of ions and
(ii)
Faraday's first law of electrolysis.
(b)
Using $\displaystyle \mathrm{E}_{\text {values }}^{\circ}$ of X and Y given below, predict which is better for coating the surface of iron to prevent corrosion and why ? Given $\displaystyle \mathrm{E}_{\mathrm{X}^{2+} / \mathrm{X}}^{\circ}=-2.36 \mathrm{~V}$, \[\begin{aligned} & \mathrm{E}_{\mathrm{Y}^{2+} / \mathrm{Y}}^{\circ}=-0.14 \mathrm{~V}, \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V} \end{aligned} \]
Marking-scheme solution
(i)
The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
(ii)
The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
(b)
'X' is better, as X has more negative electrode potential than Fe / X has more oxidation potential than Fe.
ElectrochemistryConductance of Electrolytic SolutionsApplyshort_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.