Question

Calculate the cell potential for a cell made up of the following. Ag* (aq) Li+ (aq) eAg (s) → Potential = 0.80 V Potential -3.05 V + e- + e- Li(s) [Ag+]-1.0 M and [Li+] = 0.125 M

Please show steps on how to solve this problem.

Answer 1

Lets find Eo 1st

Eo(Li+/Li(s)) = -3.05 V

Eo(Ag+/Ag(s)) = 0.80 V

As per given reaction/cell notation,

cathode is (Ag+/Ag(s))

anode is (Li+/Li(s))

Eocell = Eocathode - Eoanode

= (0.80) - (-3.05)

= 3.85 V

The reaction is:

Ag+ (aq) + Li (s) —> Li+ (aq) + Ag (s)

Number of electron being transferred in balanced reaction is 1

So, n = 1

use:

E = Eo - (2.303*RT/nF) log {[Li+]^1/[Ag+]^1}

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 {[Li+]^1/[Ag+]^1}

E = 3.85 - (0.0591/1) log (0.125^1/1.0^1)

E = 3.85-(-5.34*10^-2)

E = 3.903 V

Answer: 3.90 V