A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
Calculate the cell potential at 25 ∘C∘C under each of the following
conditions.
Part A
[Sn2+]= 1.34×10−2 MM ; [Mn2+]= 2.51 MM .
Express your answer using two significant figures.
Part B
[Sn2+]=2.51 MM ; [Mn2+]=1.34×10−2 MM .
A)
Lets find Eo 1st
from data table:
Eo(Mn2+/Mn(s)) = -1.185 V
Eo(Sn2+/Sn(s)) = -0.13 V
As per given reaction/cell notation,
cathode is (Sn2+/Sn(s))
anode is (Mn2+/Mn(s))
Eocell = Eocathode - Eoanode
= (-0.13) - (-1.185)
= 1.055 V
Number of electron being transferred in balanced reaction is
2
So, n = 2
use:
E = Eo - (2.303*RT/nF) log {[Mn2+]^1/[Sn2+]^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 {[Mn2+]^1/[Sn2+]^1}
E = 1.055 - (0.0591/2) log (2.51^1/0.0134^1)
E = 1.055-(6.719*10^-2)
E = 0.9878 V
Answer: 0.988 V
B)
Lets find Eo 1st
from data table:
Eo(Mn2+/Mn(s)) = -1.185 V
Eo(Sn2+/Sn(s)) = -0.13 V
As per given reaction/cell notation,
cathode is (Sn2+/Sn(s))
anode is (Mn2+/Mn(s))
Eocell = Eocathode - Eoanode
= (-0.13) - (-1.185)
= 1.055 V
Number of electron being transferred in balanced reaction is
2
So, n = 2
use:
E = Eo - (2.303*RT/nF) log {[Mn2+]^1/[Sn2+]^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 {[Mn2+]^1/[Sn2+]^1}
E = 1.055 - (0.0591/2) log (0.0134^1/2.51^1)
E = 1.055-(-6.719*10^-2)
E = 1.122 V
Answer: 1.12 V
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