Question

Calculate the emf of the following cell at 298 K: Al(s)/Al3+ (0.15M)//Cu2+(0.025M) /Cu(s) (Given Eo (Al3+/Al) = -1.66 V, Eo (Cu2+/Cu) = 0.34V, log 0.15 = -0.8239, log 0.025 = - 1.6020

Answer 1

Answer:

Given =

Eo (Al3+/Al) = -1.66 V,

Eo (Cu2+/Cu) = 0.34V,

log 0.15 = -0.8239,

log 0.025 = - 1.6020

to find =  EMF

solution =

we can find EMF in 2 steps.

1. find E°cell

2. final emf

E°cell = E°cathode - E°anodes

           =0.34 - (-166)

           = 2.00V

now use the nernst equation  

E°cell = E°cell - 0.059 /n log [Al⁺³]²/[Cu⁺²]³

           = 2 - 0.059 / 6 ( 2log0.15 - 3log 0.025)

           =  2- 0.59/6 ( - 16478 +4.8062 )

           =  2 - 0.0311

           =  1.9689V

the emf of the following cell at 298 K is 1.9689V