Electrochemical Cells, Cell Potentials: Using Nernst Equation

[Gaunt101]

Homework Statement

The standard reduction potentials for Ni2+ and Sn2+ are as follows:
Ni2+ + 2e− → Ni E° = −0.23 V
Sn2+ + 2e− → Sn E° = −0.14 V

Calculate the equilibrium constant at 25°C for the reaction Sn2+(aq) + Ni(s)→ Sn(s) + Ni2+(aq)

Homework Equations

Nernst Equation
E = E⁰- (0.0592/n)*log₁₀Q

The Attempt at a Solution

I thought this would be a simple problem by just plugging the relevant values into the Nernst Equation, however there is no value for E. There are two reduction potential values for the half reactions which we can calculate the total reducing potential as E⁰ = E_anode+ E_cathode = 0.14V + (-)0.23 = -0.09

This is due to the fact that you have to reverse one of the equations to change the sign so you can account for the oxidation / reduction pair. In this case, it can be seen that Sn2+ (tin) was being reduced, so I decided to reverse the equation and make it positive.

At this point I'm unsure of how to calculate the equilibrium constant because without an E value, there is no possible way of re-arranging the equation to find a suitable value. I don't think I"m missing a condition, we're not told whether the concentrations of reactants and proudcts are standard so we can't assume that E = E⁰.

Any help would be much appreciated, thank you very much!

Gaunt.

 

[Gaunt101]

Terribly sorry, I have worked this out. Just found the relationship between Gibbs Free and the equilibrium constant, rather than the reaction quotient.

Sorry!

THe answer is 1.1*10^3

 

[Borek]

No need to be sorry - we prefer those that try by themselves over those waiting for a spoon feeding

Note that you could easily solve this problem just assuming potentials of both half reactions are equal at equilibrium:

$$-0.23 + \frac {RT}{2F} ln([Ni^{2+}]) = -0.14 + \frac {RT}{2F} ln([Sn^{2+}])$$

It can be easily rearranged to the form

$$ln(\frac{[Ni^{2+}]}{[Sn^{2+}]}) = \frac {nF(0.23-0.14)}{RT}$$

or (using simplified version of the equation):

$$log_{10}(\frac{[Ni^{2+}]}{[Sn^{2+}]}) = \frac {2\times(0.23-0.14)}{0.059}$$