Lecture Dec 1, 2006 – Chapter 16 – Electrochemistry
Calculating Electrode Potentials Worksheet #2
Problem #2
Limiting/Excess Reagent Problem
Calculate the potential of a Pt electrode immersed in a solution prepared by
mixing 25.00 mL of 0.0918 M SnCl2 with 25.00 mL of 0.1568 M FeCl3
Sn+2 + 2Fe+3
Sn4+
+ 2Fe2+
You need to figure out which one (Tin or Iron) is in excess to determine which one is
involved in the half cell potential????
Sn+2
Moles have
(.025 L)(.0918M) = 0.002295 m have
Moles need (.00392 m Fe+3)*(1mSn+2/2mFe+3) = .00196 m Sn+2 need
We have more than enough of Tin so Tin is in EXCESS!!!
Fe+3
Moles have (.025 L)(.1568 M) = .00392 m have
Moles need (.002295 m Sn+2)*(2mFe+3/1mSn+2) = .00459 m Fe+3 need
We don’t have enough moles of Fe to react with Sn so Fe is limiting!
Therefore, Sn+2 is the species involved in the half cell reaction and we need to determine
how much Sn4+ is produced.
TO determine how much Sn4+ is produced, we need to use the limiting reagent moles.
(0.00293 m Fe+3)*(1 m Sn+4/2mFe+3) = 0.00196 m Sn+4
Now use the Nearnst Equation to determine the potential of the half cell
Sn+4
+
2e-
Sn+2
E0 = +0.154 V
E = E0 – 0.0592/2 (log([Sn+2]/[Sn+4]))
[Sn+2] = ((mSn+2 have – m Sn+2 need )/Total Volume) = ((0.00196-0.002295)/(.05L) =
0.0067 M Sn+2
[Sn+4] = m Sn+4 produced / Volume total = 0.00196 m / .050 L = 0.0392M Sn+4
E = E0 – 0.0592/2 (log([Sn+2]/[Sn+4]))
E = 0.154 – 0.0592/2 (log(.0067/.0392))
E = 0.177 V
In class, we also discussed the reduction of Cu+2.
How spontaneous if the reduction of Cu+2 to Cu+1 in comparison to the reduction of
Cu+2 to Cu(s)?
We can look at E (standard reduction potential) to get a GENERAL idea BUT we
need to look at deltaG to determine the relative spontaneity.
Cu+2 + eCu+2 + 2e-
Cu+1 E0 = 0.15 V
Cu(s) E0 = 0.34 V
From the reduction potentials, we can see that the reduction of Cu+2 to Cu(s) is more
spontaneous than the reduction to Cu+1.
However, delta G measures spontaneity ; we can relate delta G to E by the following
equation
deltaG = -nFE
Cu+2 + eCu+2 + 2e-
Cu+1 deltaG = -0.15F
Cu(s) deltaG = -0.68F
A NEGATIVE deltaG is a spontaneous reaction and again the reduction to Cu(s) is more
spontaneous than the reduction to Cu+1 .
And, it is more obvious from looking at the deltaG values since deltaG depends on the
mole electrons and E does not.
Therefore, the reduction to only Cu+1 will depend on the number of mole electrons
available for reduction due to the reducing agent involved.
We also talked about determining the half cell potential by taking into account other
ions in the solution. Precipitates can form in solution along with complexes.
For example, see problem #1 on “Calculating Electrode Potentials Worksheet #2”
1. Calculate the electrode potential of a Cu electrode immersed in 0.0250 M
Cu(NH3)42+and 0.128 M NH3
Cu+2 + 2e-
Cu(s) E0 = 0.34 V
E = E0 – 0.0592/2 (log(1/[Cu+2]))
In solution, Cu+2 can complex with NH3 to form Cu(NH3)42+ so we need to take into
account the formation constant for this complex.
Cu+2 + 4NH3
Cu(NH3)42+ ; β = [Cu(NH3)42+ ]/([Cu2+][NH3]4) = 5.62 x 1011
[Cu2+] =
[Cu(NH3)42+ ]/(β [NH3]4)
E = 0.34 – 0.0592/2 (log((β [NH3]4)/ [Cu(NH3)42+ ])
E = 0.34 – 0.0592/2 (log((5.62 x 1011 [0.128]4)/ [0.0250])
E = 0.0475 V
We also STARTED to talk in class about determining standard reduction potentials
for complexes.
The GENERAL Nearnst equation for the following equation is as follows:
aA + bB
cC
+
dD
E = E0 - (0.0592/n)log(([A]a[B]b)/([C]c[D]d))
E = E0 when the concentration or more correctly the activity of A, B, C, and D is
UNITY.
WE use concentration since we assume that [A] = activity of a
When we do this, we are assuming that the activity coefficient (γ) is 1 .
a = γ[A]
The activity coefficient depends on ionic strength.
For the activity coefficient to be low, the ionic strength has to be low or the ions have
to be singly charged,
REVIEW of determining ionic strength.
Ionic Strength of 0.1 M NaCl
µ = ½ {([0.1]11) + ([0.1]11)} = 0.1
When the ions are singly charged, the ionic strength equals the molarity.