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Chemistry · 2025 · 3 marks
CBSE 2025 · Region 4 · Set 1 · Q23
Calculate the cell voltage of the voltaic cell which is set up by joining following half-cells at 25^° C: Al / Al^3+(0 · 001 M ) and Ni / Ni^2+(0 · 1…
Calculate the cell voltage of the voltaic cell which is set up by joining following half-cells at $\displaystyle 25^{\circ} \mathrm{C}$ : $\displaystyle \mathrm{Al} / \mathrm{Al}^{3+}(0 \cdot 001 \mathrm{M})$ and $\displaystyle \mathrm{Ni} / \mathrm{Ni}^{2+}(0 \cdot 1 \mathrm{M})$ Given : $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\mathrm{o}}=-0.25 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\mathrm{o}}=-1.66 \mathrm{~V}$
Marking-scheme solution
2Al + $\displaystyle 3$ Ni2+ → 2Al3+ + 3Ni
E°cell=E°Ni2+/Ni - E°Al3+/Al ; E°cell=-$\displaystyle 0.25$-(-$\displaystyle 1.66$)=1.41V
n=$\displaystyle 6$
Ecell=E°cell-2.303RT log [Al3+]$\displaystyle 2$
nF [Ni2+]$\displaystyle 3$
Ecell=$\displaystyle 1.41$-$\displaystyle 0.059$ log [$\displaystyle 0.001$]$\displaystyle 2$
$\displaystyle 6$ [$\displaystyle 0.1$]$\displaystyle 3$
Ecell=$\displaystyle 1.41$-$\displaystyle 0.059$ log [$\displaystyle 10$]-$\displaystyle 6$
$\displaystyle 6$ [$\displaystyle 10$]-$\displaystyle 3$
Ecell=$\displaystyle 1.41$-$\displaystyle 0.059$ log $\displaystyle 10$-$\displaystyle 3$
Ecell=$\displaystyle 1.41$-(-$\displaystyle 0.0295$)
Ecell=$\displaystyle 1.41$+$\displaystyle 0.0295$
Ecell=1.439V/1.44V
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CBSE Class 12 Chemistry past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.