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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 3 · Q22
Calculate emf of the following cell at 298 K: Zn ( s )| Zn^2+( aq )(0.1 M ) ∥ Ag^+( aq )(0.01 M )| Ag ( s ) (Given: E_Zn^2+ / Zn^o =-0.76 V, E_Ag^+ /…
Calculate emf of the following cell at $\displaystyle 298$ K : \[\mathrm{Zn}(\mathrm{~s})\left|\mathrm{Zn}^{2+}(\mathrm{aq})(0.1 \mathrm{M}) \| \mathrm{Ag}^{+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Ag}(\mathrm{~s}) \] (Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=+0.80 \mathrm{~V},[\log 10=1]$ )
Marking-scheme solution
=
−
$\displaystyle 0.80$ – (– $\displaystyle 0.76$) = 1.56V
= $\displaystyle 1.56$ − $\displaystyle 0.0295$ log $\displaystyle 103$
= $\displaystyle 1.56$ −$\displaystyle 3$ ($\displaystyle 0.0295$)
= $\displaystyle 1.56$ − $\displaystyle 0.0885$ = $\displaystyle 1.4715$ 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.