Question

Calculate the ∆G for a cell composed of a Zn electrode in a 1.0 M Zn(NO3)3 solution and an Al electrode in a 0.1 M Al(NO3)3 solution at 298 K.

Answer 1

Lets find Eo 1st

from data table:

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

Eo(Zn2+/Zn(s)) = -0.7618 V

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

here:

cathode is (Zn2+/Zn(s))

anode is (Al3+/Al(s))

The chemical reaction taking place is

3 Zn2+(aq) + 2 Al(s) --> 3 Zn(s) + 2 Al3+(aq)

Eocell = Eocathode - Eoanode

= (-0.7618) - (-1.662)

= 0.9002 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

E = Eo - (2.303*RT/nF) log {[Al3+]^2/[Zn2+]^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/[Zn2+]^3}

E = 0.9002 - (0.0591/6) log (0.1^2/1.0^3)

E = 0.9002-(-1.971*10^-2)

E = 0.9199 V

Answer: 0.920 V