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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 2 · Q22
Calculate emf of the following cell at 298 K: Al ( s )| Al^3+( aq )(0.1 M ) ∥ Ni^2+( aq )(0.01 M )| Ni ( s ) (Given: E_Ni^2+ / Ni^o =-0.25 V, E_Al^3+…
Calculate emf of the following cell at $\displaystyle 298$ K : \[\mathrm{Al}(\mathrm{~s})\left|\mathrm{Al}^{3+}(\mathrm{aq})(0.1 \mathrm{M}) \| \mathrm{Ni}^{2+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Ni}(\mathrm{~s}) \] (Given : $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\mathrm{o}}=-0.25 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\mathrm{o}}=-1.66 \mathrm{~V},[\log 10=1]$ )
Marking-scheme solution
= (– $\displaystyle 0.25$) – (– $\displaystyle 1.66$) = 1.41V
𝐸cell = $\displaystyle 1.41$ −$\displaystyle 0.059$
𝑙𝑜𝑔
[$\displaystyle 10$−$\displaystyle 1$]$\displaystyle 2$
[$\displaystyle 10$−$\displaystyle 2$]$\displaystyle 3$
= $\displaystyle 1.41$ −
𝑙𝑜𝑔( $\displaystyle 10$+$\displaystyle 4$)
= $\displaystyle 1.41$ −
× $\displaystyle 4$ 𝑙𝑜𝑔$\displaystyle 10$
= $\displaystyle 1.41$ - $\displaystyle 0.039$
= $\displaystyle 1.371$ 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.