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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Boxxxed on April 20, 2011, 02:13:44 PM

Title: Voltaic Cell Calculations
Post by: Boxxxed on April 20, 2011, 02:13:44 PM

Question 5.  A voltaic cell is constructed using the following half-reactions:
   Cathode:       Pb²⁺ (aq) + 2 e- ---> Pb(s)
   Anode:       Pb(s) + 2 Cl-(aq) ----> PbCl₂(s) + 2 e-
   Global reaction:   Pb²⁺ + 2 Cl- ---> PbCl₂(s)

a.  If a potential of 0.142 V is measured for this cell under standard conditions, what is the standard oxidation potential (E^oox) for the anode reaction?

I can't find the oxidation potential values in the text but I am assuming this is just E cell=Cathode-Anode?

Also, anywhere online I can find these values so I can actually do the questions?

Edit:

Pb²⁺ (aq) + 2 e-  Pb(s) = -0.13

Anode = -(ECell-Cathode) = -0.272 v


b) Calculate dG^o for the cell reaction.
For this question I am not sure if I am supposed to use E cell given in the question or calculate non-standard Ecell? Concentrations aren't given until c) but G^o requires use of E^o

Title: Re: Voltaic Cell Calculations
Post by: rabolisk on April 20, 2011, 02:51:18 PM

:delta: G° and E° are independent of concentration.

Title: Re: Voltaic Cell Calculations
Post by: Boxxxed on April 20, 2011, 02:56:37 PM

Ok. I was mixing up the E and E^o.

How would I find the ksp value of PbCl2?

Ink=nE^ocell/0.0257?

Ink=(2)(-0.272)/0.0257 = 21.167

K = 1559007402

ksp = 1/k = 6.4x10^-10

This isn't right. How do I find it?

Edit: I just subbed in a wrong value.

Title: Re: Voltaic Cell Calculations
Post by: Boxxxed on April 20, 2011, 04:13:43 PM

d.  If both half-cells in the reaction initially contain 0.1 M Pb2+ (aq) and 0.1 M Cl-(aq), respectively, what potential would be measured for the cell?

ECell = 0.142 - (0.0257/2) x In(1/0.1³)

Are the substituted numbers correct? Namely the Q values?

http://img859.imageshack.us/i/pbcl2.jpg

Above is a chart summarizing some concepts. I've marked my answers but I'm not sure if they are correct as I don't fully understand this yet. Please take a look