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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 1 · Q22
Calculate emf of the following cell at 298 K: Cr ( s )| Cr^3+( aq )(0.1 M ) ∥ Fe^2+( aq )(0.01 M )| Fe ( s ) (Given: E_Cr^3+ / Cr^o =-0.74 V, E_Fe^2+…
Calculate emf of the following cell at $\displaystyle 298$ K : $\displaystyle \mathrm{Cr}(\mathrm{s})\left|\mathrm{Cr}^{3+}(\mathrm{aq})(0.1 \mathrm{M}) \| \mathrm{Fe}^{2+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Fe}(\mathrm{s})$ (Given : $\displaystyle \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\mathrm{o}}=-0.44 \mathrm{~V},[\log 10=1]$ )
Marking-scheme solution
$\displaystyle 2$ Cr(s) +3Fe2+(aq) $\displaystyle 3$ Fe(s)+ 2Cr3+(aq)
E°cell = E°cathode – E°anode
= (– $\displaystyle 0.44$) – (– $\displaystyle 0.74$) V
= $\displaystyle 0.30$ V
=
−
= $\displaystyle 0.30$ – $\displaystyle 0.0393$
= $\displaystyle 0.2606$ V or $\displaystyle 0.26$ V
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.