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 : $\displaystyle \mathrm{E}_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\mathrm{o}}=-0.25 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\mathrm{o}}=-1.66 \mathrm{~V}$

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