a voltanic cell is set up in 25 degre c Al|Al3+(0.001M) and Ni| Ni2+(0.50M) calculate cell voltage

ans -1.46v

Oxidation reaction occurs on aluminium electrode -
2Al $\to$2Al^​3+ + 6e....... E^o = 1.66 V
While reduction reaction takes place at nickel electrode -
3Ni²⁺ + 6e $\to$ 3Ni ....... E^o​ = -0.25 V
Balancer over all reaction is 
​2Al + 3Ni³⁺ $\to$2Al^​3+ + 3Ni
Thus, E^o_cell  = E^ocathode - E^o​anode
 E^o_cell   = -0.25 + 1.66 = 1.41 V
Now, using nerst equation,
E_cell  = E^o_cell  - $\frac{0.059}{n}\log \frac{[A{l}^{3+}{]}^2}{[N{i}^{2+}{]}^3}$
E_cell  = E^o_cell - $\frac{0.059}{6}\log \frac{[0.001{]}^2}{[0.5{]}^3}$

E_cell  = E^o_cell  -$0.00983 \log \frac{{10}^{-6}}{0.125}$
E_cell  = E^o_cell - 0.00983 log 8 $\times$10^-6
E_cell  = E^o_cell -0.00983 log 8 - 0.00983 log 10^-6
E_cell  = E^o_cell  - 0.008877 +  0.05898
E_cell  =  1.46011 V