Question

Consider a galvanic cell based on the following half reactions:

	E° (V)
Al3+ + 3e- → Al	-1.66
Ba2+ + 2e- → Ba	-2.90

If this cell is set up at 25°C with [Ba²⁺] = 2.50 × 10^-3M and [Al³⁺] = 4.00 × 10^-2M, the expected cell potential is

____ V

Answer 1

Lets find Eo 1st

from data table:

Eo(Ba2+/Ba(s)) = -2.90 V

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

As per given reaction/cell notation,

cathode is (Al3+/Al(s))

anode is (Ba2+/Ba(s))

Eocell = Eocathode - Eoanode

= (-1.66) - (-2.90)

= 1.24 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

use:

E = Eo - (2.303*RT/nF) log {[Ba2+]^3/[Al3+]^2}

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 {[Ba2+]^3/[Al3+]^2}

E = 1.24 - (0.0591/6) log (0.0025^3/0.04^2)

E = 1.24-(-4.937*10^-2)

E = 1.289 V

Answer: 1.29 V