Answer:
Here's what I get
Explanation:
You haven't shown your drawing, so I will assume that Metal A is the anode and Metal B is the cathode.
I will make a galvanic cell using Mg and Zn as the metals.
I selected Zn because it is common and readily available in the lab.
Zn is lower than Mg in the activity series, so Mg should be able to displace Zn from its salts
The standard reduction potentials are:
<u> E°/V </u>
Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s); -0.76
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s); -2.38
The Mg half-reaction has the more negative potential, so it will be the oxidation half-reaction.
8 and 9. Oxidation and reduction half-reactions and cell potential
<u> E°/V</u>
Oxidation: Mg(s) ⇌ Mg²⁺(aq) + 2e⁻ ; +2.38
<u>Reduction</u><u>: Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s);</u> <u>-0.76</u>
Mg(s) + Zn²⁺ (aq) ⇌ Mg²⁺(aq) + Zn(s); +1.62
The cell potential is positive, so the reaction will be spontaneous.
Mg is the anode, so it is Metal A.
Zn is the cathode, so it is Metal B.
1. The Mg|Mg²⁺ half-cell is the oxidation compartment.
2. The Zn²⁺|Zn half-cell is the reduction compartment.
3. The electrons flow from anode to cathode in the external circuit.
4. The Mg²⁺ ions flow from the Mg through the solution to the salt bridge.
5. The Zn²⁺ ions flow from the solution to the Zn.
6. NO₃⁻ ions flow from the salt bridge into the anode compartment to balance the charge of the developing Zn²⁺ ions.
7. Na⁺ ions flow from the salt bridge into the cathode compartment to replace the charge of the depleted Zn²⁺ ions.
