Question

Consider a galvanic cell based on the following half reactions:

	E° (V)
Al3+ + 3e- → Al	-1.66
Ni2+ + 2e- → Ni	-0.23

If this cell is set up at 25°C with [Ni²⁺] = 1.00 × 10^-3M and [Al³⁺] = 2.00 × 10^-2M, the expected cell potential is what?

Answer 1

Lets find Eo 1st

from data table:

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

Eo(Ni2+/Ni(s)) = -0.23 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

= (-0.23) - (-1.66)

= 1.43 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 = 1.43 - (0.0591/6) log (0.02^2/0.001^3)

E = 1.43-(5.521*10^-2)

E = 1.375 V

Answer: 1.37 V