Question

Find the cell pontential for the following volatic cell below and draw the cell diagram

Pb(s) | [Pb²⁺] (0.15 M) || Ag (s) | [Ag⁺] (0.053 M)

Given:

Pb²⁺+2e- → Pb(s) (-0.13 V)

Ag⁺+e- → Ag(s) (0.80 V)

Answer 1

from data table:

Eo(Pb2+/Pb(s)) = -0.13 V

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

As per given reaction/cell notation,

cathode is (Ag+/Ag(s))

anode is (Pb2+/Pb(s))

Eocell = Eocathode - Eoanode

= (0.80) - (-0.13)

= 0.93 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Ag+]^2/[Pb2+]^1}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/F

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Ag+]^2/[Pb2+]^1}

E = 0.93 - (0.0591/2) log (0.053^2/0.15^1)

E = 0.93-(-5.107*10^-2)

E = 0.981 V

Answer: 0.981 V