Question

Consider the cell described below at 275 K:

Sn | Sn²⁺ (0.775 M) || Pb²⁺ (0.931 M) | Pb

Given E^o_Pb²⁺→Pb = -0.131 V, E^o_Sn²⁺→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn²⁺ to have changed by 0.381 mol/L.

Tries 0/45

Answer 1

The reaction taking place is:

Sn + Pb2+ —> Sn2+ + Pb

since Sn2+ is in product, its concentration must be increasing

finally,

[Sn2+] = 0.775 + 0.381 = 1.156 M

[Pb2+] will decrease by same amount

[Pb2+] = 0.931 - 0.381 = 0.550 M

Lets find Eo 1st

Eo(Sn2+/Sn(s)) = -0.143 V

Eo(Pb2+/Pb(s)) = -0.131 V

As per given reaction/cell notation,

cathode is (Pb2+/Pb(s))

anode is (Sn2+/Sn(s))

Eocell = Eocathode - Eoanode

= (-0.131) - (-0.143)

= 0.012 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 {[Sn2+]^1/[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 {[Sn2+]^1/[Pb2+]^1}

E = 1.2*10^-2 - (0.0591/2) log (1.254^1/0.672^1)

E = 1.2*10^-2-(8.01*10^-3)

E = 3.99*10^-3 V

Answer: 3.99*10^-3 V