Question

For a spontaneous Ni, Al galvanic cell, calculate E_cell if [Al³⁺] = 2.15 M, and [Ni²⁺] = 9.99*10^-3 M.

Answer 1

from data table:

Eo(Al3+/Al(s)) = -1.662 V

Eo(Ni2+/Ni(s)) = -0.25 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ni2+/Ni(s))

anode is (Al3+/Al(s))

The chemical reaction taking place is

3 Ni2+(aq) + 2 Al(s) --> 3 Ni(s) + 2 Al3+(aq)

Eocell = Eocathode - Eoanode

= (-0.25) - (-1.662)

= 1.412 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

E = Eo - (2.303*RT/nF) log {[Al3+]^2/[Ni2+]^3}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Al3+]^2/[Ni2+]^3}

E = 1.412 - (0.0591/6) log (2.15^2/0.00999^3)

E = 1.412-(6.569*10^-2)

E = 1.346 V

Answer: 1.35 V