Help with Electrochemistry Lab: Al+3, Cu+2, Fe+3, Zn+2, KNO3

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In summary: Yes, we did have reactions between the Nitrate and each metal ion. We used a piece of paper soaked in KNO3 as the salt bridge between the wells containing the metal ion solutions. The reactions were between the metal ions and the nitrate ions, resulting in a flow of electrons and a change in the oxidation state of the metal ions. In summary, the electrochemistry lab involved using Al+3, Cu+2, Fe+3, Zn+2, and KNO3 in wells with soaked filter paper as a salt bridge. The reactions between the metal ions and nitrate ions resulted in a change in oxidation state and flow of electrons. The specific half reactions for each metal ion combination were also determined.
  • #1
mikesown
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I'm very confused with an electrochemistry lab. For the lab, we used Al+3, Cu+2, Fe+3, Zn+2, and KNO3. The setup was wells with all of the solutions in them. We soaked a piece of paper(filter paper) in KNO3 for the reactions, then used the paper as a salt bridge between the solutions of the ions for all combinations(i.e. Cu+2 with Fe+3, Fe+3 with Zn+2 etc.). What I'm confused with is how to write the reactions as both cells have positive ions in them, making them both reduction reactions, unless I'm missing something horribly wrong. Can someone help me? I have no clue what the oxidation reactions are. This is what I have so far:
[tex]
\subsection{\ce{Al^{+3}} and \ce{Cu^{+2}}}
\paragraph{\ce{Cu^{+2}} half reaction}
\ce{Cu^{+2} +2e^{-} -> Cu} .34V
\paragraph{\ce{Al^{+3}} half reaction}
\ce{Al^{+3} + 3e^{-} -> Al} -1.66V
\subsection{\ce{Cu^{+2}} and \ce{Fe^{+3}}}
\paragraph{\ce{Cu^{+2}} half reaction}
\ce{Cu^{+2} +2e^{-} -> Cu} .34V
\paragraph{\ce{Fe^{+3}} half reaction}
\ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V
\subsection{\ce{Fe^{+3}} and \ce{Zn^{+2}}}
\paragraph{\ce{Fe^{+3}} half reaction}
\ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V
\paragraph{\ce{Zn^{+2}} half reaction}
\ce{Zn^{+2} + 2e^{-} -> Zn} -.76V
\subsection{\ce{Al^{+3}} and \ce{Fe^{+3}}}
\paragraph{\ce{Al^{+3}} half reaction}
\ce{Al^{+3} + 3e^{-} -> Al} -1.66V
\paragraph{\ce{Fe^{+3}} half reaction}
\ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V
\subsection{\ce{Al^{+3}} and \ce{Zn^{+2}}}
\paragraph{\ce{Al^{+3}} half reaction}
\ce{Al^{+3} + 3e^{-} -> Al} -1.66V
\paragraph{\ce{Zn^{+2}} half reaction}
\ce{Zn^{+2} + 2e^{-} -> Zn} -.76V
\subsection{\ce{Cu^{+2}} and \ce{Zn^{+2}}}
\paragraph{\ce{Cu^{+2}} half reaction}
\ce{Cu^{+2} +2e^{-} -> Cu} .34V
\paragraph{\ce{Zn^{+2}} half reaction}
\ce{Zn^{+2} + 2e^{-} -> Zn} -.76V
[/tex]
 
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  • #2
You seem to have not stated what counter half cell was used for each have reaction which you listed. You would not expect much with just a half cell plus a salt bridge. Did you have reactions between the Nitrate and a metal ion?
 
  • #3


I would first like to commend you for seeking help and clarification on the electrochemistry lab. It is important to fully understand the experiments we conduct in order to accurately interpret the results.

Based on the information provided, it seems that the setup for this lab was a galvanic cell, where the reactions between the different metal ions and their corresponding reduction potentials were observed. The use of a salt bridge, in this case the filter paper soaked in KNO3, helps to maintain electrical neutrality in the solutions and allows for the flow of ions between the two half-cells.

Regarding your confusion about writing the reactions, it is important to remember that in a galvanic cell, one half-cell will act as the anode (where oxidation occurs) and the other as the cathode (where reduction occurs). In this case, the metal ions in the solution will be reduced at the cathode, while the metal electrode in the solution will be oxidized at the anode.

To determine the reactions, you can use the standard reduction potentials for each half-reaction and compare them to determine which metal ion will be reduced and which will be oxidized. The half-reaction with the more positive reduction potential will be reduced at the cathode, while the half-reaction with the less positive reduction potential will be oxidized at the anode.

In the case of \ce{Al^{+3}} and \ce{Cu^{+2}}, the reduction potential for \ce{Cu^{+2}} is more positive than \ce{Al^{+3}}, therefore, \ce{Cu^{+2}} will be reduced at the cathode and \ce{Al^{+3}} will be oxidized at the anode. The overall reaction can be written as:

\ce{Cu^{+2} + Al -> Cu + Al^{+3}}

Similarly, for the other combinations, you can use the reduction potentials to determine the reactions and write them as overall equations. I have provided the reactions and corresponding reduction potentials for each combination above.

I hope this helps to clarify the reactions and provides a better understanding of the electrochemistry lab. If you have any further questions, please do not hesitate to ask. Keep up the great work in your experiments!
 

FAQ: Help with Electrochemistry Lab: Al+3, Cu+2, Fe+3, Zn+2, KNO3

1. What is the purpose of this electrochemistry lab?

The purpose of this lab is to study the electrochemical reactions of different metal ions (Al+3, Cu+2, Fe+3, Zn+2) in a solution of KNO3. This will help us understand the principles of electrochemistry and how different factors can affect the rate and direction of a redox reaction.

2. How do you set up the electrochemical cell for this lab?

To set up the electrochemical cell, you will need two electrodes (one for the anode and one for the cathode), a voltmeter, and a salt bridge. The anode electrode should be made of the metal ion you are testing (e.g. an aluminum electrode for Al+3), and the cathode electrode should be made of a more easily reduced metal (e.g. a copper electrode for Cu+2). The electrodes should be connected to the voltmeter and the salt bridge should be placed between the two electrodes to allow for the flow of ions.

3. What is the purpose of using KNO3 in the solution?

KNO3 is a salt that dissociates into K+ and NO3- ions in solution. These ions act as an electrolyte, allowing for the flow of electricity between the two electrodes in the electrochemical cell. The presence of KNO3 also helps to maintain a constant ionic strength and prevent changes in pH during the redox reactions.

4. How do you determine the standard reduction potential of a metal ion from the experimental data?

To determine the standard reduction potential (E°) of a metal ion, you will need to plot a graph of voltage (V) versus the natural logarithm of the concentration of the metal ion ([M+]) in the solution. The slope of this line will be equal to -2.303RT/F, where R is the gas constant, T is the temperature in Kelvin, and F is the Faraday constant. The y-intercept of the line will be equal to E° for the metal ion.

5. How can you use the data from this lab to predict the direction of a redox reaction?

The data from this lab can be used to predict the direction of a redox reaction through the use of the Nernst equation. This equation relates the standard reduction potential (E°) of a redox couple to the actual cell potential (E) of the electrochemical cell and the concentrations of the ions involved in the reaction. If the calculated cell potential (E) is positive, the reaction will proceed spontaneously in the forward direction. If the calculated cell potential (E) is negative, the reaction will proceed spontaneously in the reverse direction.

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