Question

A voltaic cell is constructed using a Sn electrode in .50 M SnSO4 and a Cd electrode in .01M CdSO4 solution. Determine the following:

a) Overall reaction

b) Metal oxidized

c) Metal Reduced

d) E0 cell

e) E cell

Answer 1

a)

Lets find Eo 1st

from data table:

Eo(Cd2+/Cd(s)) = -0.4 V

Eo(Sn2+/Sn(s)) = -0.13 V

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

here:

cathode is (Sn2+/Sn(s))

anode is (Cd2+/Cd(s))

The chemical reaction taking place is

Sn2+(aq) + Cd(s) --> Sn(s) + Cd2+(aq)

b)

Metal oxidised is Cd

c)

metal reduced is Sn

d)

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.4)

= 0.27 V

e)

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

E = Eo - (2.303*RT/nF) log {[Cd2+]^1/[Sn2+]^1}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cd2+]^1/[Sn2+]^1}

E = 0.27 - (0.0591/2) log (0.01^1/0.5^1)

E = 0.27-(-5.023*10^-2)

E = 0.32 V