CBSE 2023 · Region 5 · Set 2 · Q24 · 2 marks
(a)What should be the signs (positive/negative) for $\displaystyle \mathrm{E}^{\circ}{ }_{\text {Cell }}$ and $\displaystyle \Delta \mathrm{G}^{\circ}$ for a spontaneous redox reaction occurring under standard conditions?(ii)State Faraday's first law of electrolysis.Calculate the emf of the following cell at $\displaystyle 298$ K: $\displaystyle \mathrm{Fe}_{(\mathrm{s})}\left|\mathrm{Fe}^{2+}(0.01 \mathrm{M})\right|\left|\mathrm{H}^{+}{ }_{(1 \mathrm{M})}\right| \mathrm{H}_{2(\mathrm{~g})}(1$ bar $\displaystyle ), \mathrm{Pt}_{(\mathrm{s})}$ Given $\displaystyle \mathrm{E}^{\circ}{ }_{\text {Cell }}=0.44 \mathrm{~V}$.
(a)
What should be the signs (positive/negative) for $\displaystyle \mathrm{E}^{\circ}{ }_{\text {Cell }}$ and $\displaystyle \Delta \mathrm{G}^{\circ}$ for a spontaneous redox reaction occurring under standard conditions?
(ii)
State Faraday's first law of electrolysis.
Calculate the emf of the following cell at $\displaystyle 298$ K: $\displaystyle \mathrm{Fe}_{(\mathrm{s})}\left|\mathrm{Fe}^{2+}(0.01 \mathrm{M})\right|\left|\mathrm{H}^{+}{ }_{(1 \mathrm{M})}\right| \mathrm{H}_{2(\mathrm{~g})}(1$ bar $\displaystyle ), \mathrm{Pt}_{(\mathrm{s})}$ Given $\displaystyle \mathrm{E}^{\circ}{ }_{\text {Cell }}=0.44 \mathrm{~V}$.
Marking-scheme solution
(a) (i) \(\displaystyle \mathrm{E}^{0}{ }_{\text {cell }}=+\mathrm{ve} \quad \& \Delta \mathrm{G}^{0}=-\mathrm{ve}\)
(ii) It states that the mass of a substance deposited /liberated at the electrodes is directly proportional to the charge/quantity of electricity passed through the electrolyte.
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.