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 : $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\mathrm{o}}=-0.25 \mathrm{~V}$ \[\begin{aligned} & \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\mathrm{o}}=-1.66 \mathrm{~V} \\ & \log 5=0.6990 \end{aligned} \]

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