using standard potentials given in the appendices calculate the standard cell potentials and the equilibrium constants for the following reactions.
Zn(s)+( Fe^2+(aq)) ---> (Zn^2+(aq))+ Fe(s)
If anyone can show work, then I can do the rest of my homework by looking at this example.
Thank you!!!!!
Cathode (Reduction)
Half-Reaction Zn2+(aq) + 2e- -> Zn(s)
Standard Potential E° (volts)= -0.76
Cathode (Reduction)
Half-Reaction Fe2+(aq) + 2e- -> Fe(s)
Standard Potential E° (volts)= -0.41
Both can't be the cathode.
OK. Zn --> Zn^2+ is the half reaction in the problem. What you looked up is the standard reduction potential but this is an oxidation; therefore, reverse the sign. The E value for the half reaction as written is +0.76 v. The other is a reduction and it has the potential you have indicated. Add the oxidation half and the reduction half to obtain the total rxn.
Ecell as written is 0.76 + (-0.41 = ?
How do you calculate the equilibrium constants for these reactions? This was the second part of the question!
Thank you!
To calculate the standard cell potential, you can use the Nernst equation:
E°cell = E°cathode - E°anode
First, let's identify the half-reactions occurring at each electrode:
At the cathode:
Fe^2+(aq) + 2e^- → Fe(s) (reduction)
At the anode:
Zn(s) → Zn^2+(aq) + 2e^- (oxidation)
Now, we need to find the standard reduction potentials for these half-reactions. These values can be found in the appendices of your textbook or a reliable online resource.
From the Appendices, we find:
E°Fe^2+/Fe = +0.44 V
E°Zn^2+/Zn = -0.76 V
Using the Nernst equation, we can calculate the standard cell potential (E°cell) for the given reaction:
E°cell = E°cathode - E°anode
= (+0.44 V) - (-0.76 V)
= +1.20 V
The positive value indicates that the reaction is spontaneous in the forward direction at standard conditions.
To find the equilibrium constant (K) at standard conditions, you can use the equation:
E°cell = (0.0592 V/n) * log(K)
Rearranging the equation gives:
K = 10^((E°cell) / (0.0592 V/n))
Given the value of E°cell (+1.20 V), and considering that the half-reactions involve the transfer of two electrons (n = 2), we can calculate the equilibrium constant (K):
K = 10^((+1.20 V) / (0.0592 V/2))
= 10^20.27
≈ 1.64 * 10^20
So, the equilibrium constant for the given reaction is approximately 1.64 * 10^20.
Please note that if the reaction conditions are not at standard conditions, you will need to use a modified form of the Nernst equation to calculate the cell potential and equilibrium constant.
Look up and post the reduction reaction and Eo for each half cell so we'll be using the same numbers.
Sorry, I forgot about that.
dGo = -nFEo