# Reduction Problem in a cell containing a standard hydrogen electrode and Pd(s)|Pd2+

Submitted by dasergeo on Mon, 03/31/2008 - 21:13

concentratioin of Pd2+ in solution using a cell with SHE as one half cell and Pd(s)|Pd2+ as the other having a potential of 0.652 volts when black lead was connected to SHE
(Eored for Pd2+(aq) + 2e- --> Pd(s)      = 0.987 volts)

Now I assume I use Eocell=cathode-anode.  I have tried many ways of setting it up to find the answer but I keep getting it wrong.  Any help would be appreciated.  The answer is [Pd2+]=4.8x10-12M

You need to use the Nernst Equation since the reaction is not standard (Standard reactions are when all concentrations are 1M and gases have a pressure of 1 atm).  The fact that they are asking for the concentration of the Pd2+ indicates that the reaction is not at standard conditions.

First of all the total cell potential is the sum of the potentials of the half cells, not the difference.
E cell    =      cathode +   anode.
I assume that SHE means Standard Hydrogen Electrode which has an Eo = 0

That means the E cell =  E Pt  +   E SHE               but E SHE =0           so
E cell =  E Pt

However, the value you are giving for the Pt cell potential is for the Standard Pd cell potential (concentration of Pd2+ would be 1M).  Cell potentials will change as the concentration of the reactants and products change.  The Nernst Equation allows for these changes in concentration to be considered.

E   =   Eo    - ( .0592 /n ) x log Q  where n= number of electrons transferred and Q is the reaction quotient.  Q has the same form as the expression for the equilibrium constant.

In this case    E  =  Eo  - (.0592/2) x log ([Pd(s)]/[Pd2+])
However, since Pd metal is a solid, it is omitted from the expression for Q leaving:

E  =  Eo  - (.0592/2) x log (1 /[Pd2+])

We are told the actual cell potential at non-standard conditions is .652 V and we know the Standard Cell potential is .987 V.  If we plug these values in we can solve for [Pt2+]

.652 = .987   - ( .0592/2) x log 1/[Pd2+]
Rearranging yields
(.652 - .987) x 2/.0592 = -log (1/[Pd2+]

But log (1/[Pd2+])  =  log 1  - log [Pd2+] and log 1 = 0, so

(.652 - .987) x 2/.0592 =   log [Pd2+]

Taking the antilog of both sides yields:
[Pd2+]  = 4.81 x 10-12