Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:


With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

I hope it helps!
I think it is <span>Symbiosis. But I dont know how to explain it</span>
You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results
Sulphate ion is used but fornyour kind information there ain't no place