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

Click For Summary
SUMMARY

The discussion focuses on an electrochemistry lab involving the ions Al3+, Cu2+, Fe3+, Zn2+, and KNO3. Participants utilized filter paper soaked in KNO3 as a salt bridge to facilitate reactions between various ion combinations. The confusion arises from the identification of oxidation and reduction reactions, particularly since both cells contain positive ions. The half-reactions for each ion are provided, along with their standard reduction potentials, which are crucial for determining the direction of electron flow.

PREREQUISITES
  • Understanding of electrochemical cells and half-reactions
  • Knowledge of standard reduction potentials
  • Familiarity with salt bridge functionality in electrochemistry
  • Basic skills in writing and balancing chemical equations
NEXT STEPS
  • Study the Nernst equation for calculating cell potentials under non-standard conditions
  • Explore the concept of galvanic cells and their applications in electrochemistry
  • Learn about the electrochemical series and its role in predicting reaction spontaneity
  • Investigate the effects of concentration changes on cell potential
USEFUL FOR

This discussion is beneficial for chemistry students, educators, and laboratory technicians involved in electrochemistry experiments, particularly those focusing on redox reactions and cell potential calculations.

mikesown
Messages
16
Reaction score
0
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:
<br /> \subsection{\ce{Al^{+3}} and \ce{Cu^{+2}}}<br /> \paragraph{\ce{Cu^{+2}} half reaction}<br /> \ce{Cu^{+2} +2e^{-} -&gt; Cu} .34V<br /> \paragraph{\ce{Al^{+3}} half reaction}<br /> \ce{Al^{+3} + 3e^{-} -&gt; Al} -1.66V <br /> \subsection{\ce{Cu^{+2}} and \ce{Fe^{+3}}}<br /> \paragraph{\ce{Cu^{+2}} half reaction}<br /> \ce{Cu^{+2} +2e^{-} -&gt; Cu} .34V<br /> \paragraph{\ce{Fe^{+3}} half reaction}<br /> \ce{Fe^{+3} + e^{-} -&gt; Fe^{+2}} .77V<br /> \subsection{\ce{Fe^{+3}} and \ce{Zn^{+2}}}<br /> \paragraph{\ce{Fe^{+3}} half reaction}<br /> \ce{Fe^{+3} + e^{-} -&gt; Fe^{+2}} .77V<br /> \paragraph{\ce{Zn^{+2}} half reaction}<br /> \ce{Zn^{+2} + 2e^{-} -&gt; Zn} -.76V<br /> \subsection{\ce{Al^{+3}} and \ce{Fe^{+3}}}<br /> \paragraph{\ce{Al^{+3}} half reaction}<br /> \ce{Al^{+3} + 3e^{-} -&gt; Al} -1.66V <br /> \paragraph{\ce{Fe^{+3}} half reaction}<br /> \ce{Fe^{+3} + e^{-} -&gt; Fe^{+2}} .77V<br /> \subsection{\ce{Al^{+3}} and \ce{Zn^{+2}}}<br /> \paragraph{\ce{Al^{+3}} half reaction}<br /> \ce{Al^{+3} + 3e^{-} -&gt; Al} -1.66V <br /> \paragraph{\ce{Zn^{+2}} half reaction}<br /> \ce{Zn^{+2} + 2e^{-} -&gt; Zn} -.76V<br /> \subsection{\ce{Cu^{+2}} and \ce{Zn^{+2}}}<br /> \paragraph{\ce{Cu^{+2}} half reaction}<br /> \ce{Cu^{+2} +2e^{-} -&gt; Cu} .34V<br /> \paragraph{\ce{Zn^{+2}} half reaction}<br /> \ce{Zn^{+2} + 2e^{-} -&gt; Zn} -.76V<br />
 
Physics news on Phys.org
Did you try to measure an EMF on each of these cells you created? If so, what metal was used for the electrodes in each case?
 

Similar threads

Replies
1
Views
3K
Replies
13
Views
14K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
12K
  • · Replies 2 ·
Replies
2
Views
32K
Replies
21
Views
6K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
9K