[elgoog]

I collected this data from my experiment:

Standard oxidation potential for zinc:
Zn Zn²+ + 2e- = +0.76 volts.

Standard reduction potential for copper:
Cu₂+ + 2e- Cu = +0.34 volts

I calculated the equilibrium constant and came up with this:
E⁰= .896v
log(K)=ne/0.0591
log(K)=2(0.877)/0.0591
log(K)=29.68
K=10_29.68

Is the rate constant answer correct?

[Borek]

E⁰= .896v

Where did this value come from?

Is the rate constant answer correct?

Rate, or equilibrium?

[elgoog]

I have no idea where I got .896 volts from. I should've subracted 0.34 volts from 0.76 volts to get an E⁰ of 0.42 volts.

How does this look:

E⁰= 0.42 volts
log(K)=ne/0.0591
log(K)=2(0.42)/0.0591
log(K)=14.21
K=10^14.21
K=1.62 x 10¹⁴

I am trying to calculate the equilibrium constant for a redox reaction.

[Borek]

I have no idea where I got .896 volts from. I should've subracted 0.34 volts from 0.76 volts to get an E⁰ of 0.42 volts.

You are getting closer, but there is still a problem - this time with your initial data. Standard potential for the redox reaction for Zn/Zn²⁺ is not 0.76V.

Or, to put it differently - you are mixing different approaches to calculate the cell potential, and you are getting the wrong final answer.

My advice: never use oxidation potentials. When you have a half cell in the solution it will react in both directions at exactly the same potential, it will not automagically switch the sign just because the reaction is reversed.

[elgoog]

Is this better?

The standard reduction potential for zinc is: Zn2+  + 2e- Zn = -0.76volts             
The standard reduction potential for copper is: Cu2+  + 2e- Cu = 0.34volts
         
   log(K) = nE⁰ ÷ 0.0591
   log(K) = 2(-0.42) ÷ 0.0591
   log(K) = -14.21
   K = 10^-14.21
   K = 6 x 10^-15

[Borek]

You are juggling numbers and I have no idea what you got and from where.

Let's concentrate on the potential of the Cu/Zn cell, assuming standard conditions. Zinc reacts at -0.76V, copper reacts at 0.34V. What is the cell voltage?

[elgoog]

According to this website: http://www.chem.purdue.edu/gchelp/howtosolveit/Electrochem/Electrochemical_Cell_Potentials.htm

The standard potentials for the reduction half reaction are:

Zn: -0.76 volts
Cu: 0.34 volts

The sign on Zn needs to be reversed because it is the oxidation reaction so it becomes 0.76 volts instead of -0.76 volts.

When I add 0.76 plus 0.34 together, I get 1.1 volts.

Now I'm going to use the equation from my textbook located here: https://drive.google.com/a/wgu.edu/file/d/0B1DPkJt51Dg4Q3RaMHljVXJLRHc/view

       
   log(K) = nE⁰ ÷ 0.0591
   log(K) = 2(1.1) ÷ 0.0591
   log(K) = 37.22
   K = 10^37.22
   K = 1.66 x 10³⁷

Does that look right now?

[Borek]

Looks OK now (unless I am missing something as well).

Please note the convention of reversing the sign because the reaction is reversed is more confusing than helping. Just because you reverse the reaction doesn't mean it takes place at different potential - at standard conditions Zn get both reduced and oxidized at -0.76 V (vs SHE).

To find the cell voltage all you need is to find the difference between potentials subtracting the lower number from the higher number - you are calculating the DISTANCE between these two potentials. If one of them is negative, because of the subtraction sign will be changed automatically. In the case of Cu/Zn system

[tex]E_0 = E_{higher} - E_{lower} = 0.34 - (-0.76) = 1.11V[/tex]

[elgoog]

Oh I see, that makes sense to me. The sign will be reversed when you subtract it anyway. Thanks a lot for your help.