Write the half cell reactions of the following?

1) Cu/Cu^2+(0.1M)//Ag^+(0.1M)/Ag 2) Zn/Zn^2+ (0.1M)//Cu^2+(0.1M)/Cu 3) Zn/Zn^2+(0.1M)//Ag^+(0.1M)/Ag 4) Zn/Zn(NH3)4^2+*0.5M)//Ag^+(0.1M)/Ag 5) Zn^2+(aq)+4NH3(aq) <--> ...show more

Also, around what range of numerical results would be expected of their cell potentials, polarities, Kfs, Keqs, and E cell values would be expected here?

You haven't asked a question here but more like a book of questions. Do you just want to "compare" answers or do you have a real problem with understanding a particular part? Clarify what you don't know and explain in detail and we can help you through it. I just don't feel like writing a book today.

I want to compare answers. I recently did an experiment on Galvanic Cells, and I obtained readings from a volt meter, and I was wondering if there were specific results in which my results must come into close range.

My cell potentials listed in order from the question are -0.12, -0.13, 0.86, 0.44, and 0.56. I had problems working the machine and I added that as a source of error.

Also for my half reactions, I got these answers from the first 3 in order from the question listed above,

Cu-->Cu^2+
Zn^2+-->Zn
Ag^+-->Ag
Someone was explaining to me on how to write half reactions over and over again and I am still confused, and I know my answers are definitely wrong.

I just realized my question was cut off, sorry:


Write the half cell reactions of the

1) Cu/Cu^2+(0.1M)//Ag^+(0.1M)/Ag
2) Zn/Zn^2+ (0.1M)//Cu^2+(0.1M)/Cu
3) Zn/Zn^2+(0.1M)//Ag^+(0.1M)/Ag
4) Zn/Zn(NH3)4^2+*0.5M)//Ag^+(0.1M)/Ag
5) Zn^2+(aq)+4NH3(aq) <--> Zn(NH3)4^2+(aq)
6) Ag/AgCl//Ag^+(0.1M)/Ag

Also around what range of numerical results would be expected of their cell potentials, polarities, Kfs, Keqs, and E cell values would be expected here?

For #2.

The half rxn are
Zn ==> Zn^2+ + 2e and
Cu^2+ + 2e == Cu

To find Ehalf cell for Zn use
E = Eocell - (0.05916/n)*log[1/(Zn^2+)]. Plug in the (Zn^2+), calculate Ehalf cell (using the reduction potential you find in your standard potential table), then reverse the sign and add it to the potential you calculate for Cu half cell. That will give you the theoretical Ecell at the concns you have listed for Cu^2+ and Zn^2+. It will also give you the sign of Ecell.
Use Ecell calculated to arrive at Keq or Kf from nFE = RT*lnK

To determine the half-cell reactions for the given cases, we need to follow a few steps. Here's the process:

Step 1: Identify the species being oxidized and reduced in the half-cell reactions.
- The species being oxidized undergoes oxidation, resulting in the loss of electrons.
- The species being reduced undergoes reduction, resulting in the gain of electrons.

Step 2: Write the balanced chemical equation for the redox reactions.
- Ensure that the number of atoms and charges are balanced on both sides of the equation.

Step 3: Determine the half-cell reactions.
- The half-cell reactions can be written by just focusing on the oxidation and reduction half-reactions.

Now, let's apply this process to the given cases:

1) Cu/Cu^2+(0.1M)//Ag^+(0.1M)/Ag
- The species being oxidized is Cu (Cu → Cu^2+ + 2e^-).
- The species being reduced is Ag^+ (Ag^+ + e^- → Ag).

2) Zn/Zn^2+(0.1M)//Cu^2+(0.1M)/Cu
- The species being oxidized is Zn (Zn → Zn^2+ + 2e^-).
- The species being reduced is Cu^2+ (Cu^2+ + 2e^- → Cu).

3) Zn/Zn^2+(0.1M)//Ag^+(0.1M)/Ag
- The species being oxidized is Zn (Zn → Zn^2+ + 2e^-).
- The species being reduced is Ag^+ (Ag^+ + e^- → Ag).

4) Zn/Zn(NH3)4^2+*0.5M)//Ag^+(0.1M)/Ag
- The species being oxidized is Zn (Zn → Zn(NH3)4^2+ + 2e^-).
- The species being reduced is Ag^+ (Ag^+ + e^- → Ag).

5) Zn^2+(aq) + 4NH3(aq) ⇌ ...

Unfortunately, the information provided for the fifth case is incomplete, as there is no complete equation or reactant/product given. Without further data, it is not possible to determine the half-cell reactions.

Regarding the expected numerical results in terms of cell potentials (Ecell), polarities, formation constants (Kf), and equilibrium constants (Keq), they depend on the specific reactions and concentrations involved. Without specific information, it is difficult to provide an exact range. However, generally, more positive cell potentials indicate a greater tendency for the reaction to occur spontaneously, and positive Ecell values imply cell reactions are feasible.

The polarity of the cell depends on the sign of the cell potential. Positive cell potentials indicate a spontaneous, positive-to-negative flow of electrons, representing a galvanic or voltaic cell. Negative cell potentials imply a non-spontaneous reaction and require an external power source, representing an electrolytic cell.

Formation constants (Kf) represent the stability of complex ions, where a higher Kf value indicates greater stability.

Equilibrium constants (Keq) quantify the position of a reversible reaction, with larger Keq values indicating a more favorable equilibrium position.

Remember, for more precise numerical values, you would need specific concentration values and the full reaction equations.