A voltaic cell consists of an AlAl3 halfcell and a AgAg half

A voltaic cell consists of an Al/Al³⁺ half-cell and a Ag/Ag⁺ half-cell. Calculate E_cell when [Ag⁺] = 0.00160 M and [Al³⁺] = 0.9040 M.
You should use the reduction potentials for Al³⁺ is -1.66 V and for Ag⁺ is 0.80 V.

Solution

Al + 3Ag+ --->Ag(s) +Al3+

so Eo=0.8+1.66

=2.46 V

so,

E=Eo-0.059/3 log([Al+]/[Ag3+]^3)

=2.46 -(0.059/3) * log(0.904/(0.0016^3))

=2.2959 V