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

[mikesown]

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:
$$\subsection{\ce{Al^{+3}} and \ce{Cu^{+2}}} \paragraph{\ce{Cu^{+2}} half reaction} \ce{Cu^{+2} +2e^{-} -> Cu} .34V \paragraph{\ce{Al^{+3}} half reaction} \ce{Al^{+3} + 3e^{-} -> Al} -1.66V \subsection{\ce{Cu^{+2}} and \ce{Fe^{+3}}} \paragraph{\ce{Cu^{+2}} half reaction} \ce{Cu^{+2} +2e^{-} -> Cu} .34V \paragraph{\ce{Fe^{+3}} half reaction} \ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V \subsection{\ce{Fe^{+3}} and \ce{Zn^{+2}}} \paragraph{\ce{Fe^{+3}} half reaction} \ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V \paragraph{\ce{Zn^{+2}} half reaction} \ce{Zn^{+2} + 2e^{-} -> Zn} -.76V \subsection{\ce{Al^{+3}} and \ce{Fe^{+3}}} \paragraph{\ce{Al^{+3}} half reaction} \ce{Al^{+3} + 3e^{-} -> Al} -1.66V \paragraph{\ce{Fe^{+3}} half reaction} \ce{Fe^{+3} + e^{-} -> Fe^{+2}} .77V \subsection{\ce{Al^{+3}} and \ce{Zn^{+2}}} \paragraph{\ce{Al^{+3}} half reaction} \ce{Al^{+3} + 3e^{-} -> Al} -1.66V \paragraph{\ce{Zn^{+2}} half reaction} \ce{Zn^{+2} + 2e^{-} -> Zn} -.76V \subsection{\ce{Cu^{+2}} and \ce{Zn^{+2}}} \paragraph{\ce{Cu^{+2}} half reaction} \ce{Cu^{+2} +2e^{-} -> Cu} .34V \paragraph{\ce{Zn^{+2}} half reaction} \ce{Zn^{+2} + 2e^{-} -> Zn} -.76V$$

 

[chemisttree]

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?

 

[Blueshift5]

I can understand your confusion with the electrochemistry lab. It can be a complex topic to grasp, but I will try my best to explain it to you.

Firstly, it is important to understand that electrochemistry deals with the transfer of electrons in chemical reactions. In your lab, you were observing the reactions between positive ions of different metals. These reactions involve the transfer of electrons from one metal to another, resulting in the formation of new compounds.

To write the reactions, we need to identify the oxidation and reduction half reactions. In an oxidation reaction, a substance loses electrons, while in a reduction reaction, a substance gains electrons. In your case, the metals are losing electrons to form positive ions, so they are undergoing oxidation reactions.

Let's take the first combination of \ce{Al^{+3}} and \ce{Cu^{+2}}. The half reactions would be:

\paragraph{\ce{Cu^{+2}} oxidation half reaction}
\ce{Cu^{+2} -> Cu + 2e^{-}}

\paragraph{\ce{Al^{+3}} reduction half reaction}
\ce{Al^{+3} + 3e^{-} -> Al}

As you can see, the electrons lost by copper are gained by aluminum, resulting in the formation of solid copper and aluminum ions in solution. The overall reaction would be:

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

Similarly, for the other combinations, we can write the half reactions and overall reactions as follows:

\subsection{\ce{Cu^{+2}} and \ce{Fe^{+3}}}
\paragraph{\ce{Cu^{+2}} oxidation half reaction}
\ce{Cu^{+2} -> Cu + 2e^{-}}

\paragraph{\ce{Fe^{+3}} reduction half reaction}
\ce{Fe^{+3} + e^{-} -> Fe^{+2}}

Overall reaction: \ce{Cu^{+2} + Fe^{+3} -> Cu + Fe^{+2}}

\subsection{\ce{Fe^{+3}} and \ce{Zn^{+2}}}
\paragraph{\ce{Fe^{+3}} reduction half reaction}
\ce{Fe^{+3} + e^{-} -> Fe^{+2}}

\paragraph{\ce{Zn^{+2}} oxidation half reaction}
\ce{Zn^{+2