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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 2 · Set 1 · Q23
Calculate emf of the following cell at 298 K: Sn | Sn^2+(0.001 M )|| H^+(0.01 M )| H_2( g ) (1 bar ) ∣ Pt_( s ) Given: E^0_Sn^2+ / Sn =-0.14 V,…
Calculate emf of the following cell at $\displaystyle 298$ K : $\displaystyle \mathrm{Sn}\left|\mathrm{Sn}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.01 \mathrm{M})\right| \mathrm{H}_{2(\mathrm{~g})}(1$ bar $\displaystyle ) \mid \mathrm{Pt}_{(\mathrm{s})}$ Given : $\displaystyle \mathrm{E}^{0}{ }_{\mathrm{Sn}^{2+} / \mathrm{Sn}}=-0.14 \mathrm{~V}$, \[\mathrm{E}^{\mathrm{o}}{ }_{\mathrm{H}^{+} / \mathrm{H}_{2}}=0.00 \mathrm{~V}[\log 10=1] \]
Marking-scheme solution
Sn(s) (Sn2+(0.001M) ((H+(0.01M) (H2(g)($\displaystyle 1$ bar) (Pt(s)
E°cell = $\displaystyle 0$ − (− $\displaystyle 0.14$ V) = $\displaystyle 0.14$ V
𝐸cell = E°cell − $\displaystyle 0.059$
log
[Sn2+]
[𝐻+] $\displaystyle 2$
= $\displaystyle 0.14$ V -
log
= $\displaystyle 0.14$ V - $\displaystyle 0.059$
log $\displaystyle 10$
=$\displaystyle 0.14$ V − $\displaystyle 0.0295$ V
= $\displaystyle 0.1105$ 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.