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Chemistry · 2025 · 3 marks
CBSE 2025 · Region 4 · Set 3 · Q28
Calculate the cell voltage of the voltaic cell which is set up by joining the following half-cells at 25°C. Al / Al^3+(0 · 002 M ) and Ni / Ni^2+(0 ·…
Calculate the cell voltage of the voltaic cell which is set up by joining the following half-cells at $\displaystyle 25$°C. \[\mathrm{Al} / \mathrm{Al}^{3+}(0 \cdot 002 \mathrm{M}) \text { and } \mathrm{Ni} / \mathrm{Ni}^{2+}(0 \cdot 002 \mathrm{M}) \] Given : $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\mathrm{o}}=-0.25 \mathrm{~V}$ \[\begin{aligned} & \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\mathrm{o}}=-1.66 \mathrm{~V} \\ & \log 5=0.6990 \end{aligned} \]
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 [$\displaystyle 0.002$]$\displaystyle 2$
$\displaystyle 6$ [$\displaystyle 0.002$]$\displaystyle 3$
Ecell =$\displaystyle 1.41$-$\displaystyle 0.059$ log [2x10-$\displaystyle 3$]$\displaystyle 2$
$\displaystyle 6$ [2x10-$\displaystyle 3$]$\displaystyle 3$
Ecell =$\displaystyle 1.41$-$\displaystyle 0.059$ log (5x102)
Ecell =$\displaystyle 1.41$-$\displaystyle 0.059$ ($\displaystyle 2.6990$)
Ecell =$\displaystyle 1.41$-$\displaystyle 0.0265$
Ecell =$\displaystyle 1.3835$ V
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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.