i had 2 question left to finish but i cant really undersdand how to do it

9)design an elextrochemical cell using Pb(s) and Mn(s) and theire solutions and answer the following questions:
i answerd a-i all that is left is j which is:
indicate the substance oxidise,substance reduce, oxidizing agent, and reducing agent
if i say Mn is substance reduction and Pb is also substance reduction would that be correct?

next question is 11) for the electrolysis of an aqueous solution of AuCl3(aq)
i did a-c all that is left is d) state the products of the electrolysis and i have no ideo what to do here
thanks in advance

No, the material you name as the reducing agent can't be two different materials.

The substance oxidized is the one that loses electrons and the substance reduced is the one that gains electrons.
The oxidizing agent is the one reduced. The reducing agent is the one oxidized.

Au is plated out at the cathode and Cl2 is produced at the anode

For Question 9:

To design an electrochemical cell using Pb(s) and Mn(s) and their solutions, you'll need to follow these steps:

1. Identify the half-reactions: First, you need to determine the half-reactions for each electrode. In this case, the half-reaction for lead (Pb) can be written as Pb(s) → Pb2+(aq) + 2e-. The half-reaction for manganese (Mn) can be written as Mn(s) → Mn2+(aq) + 2e-.

2. Assign electrodes: Choose the corresponding electrode for each half-reaction. For the oxidation reaction, where lead is losing electrons, you would use a lead electrode (Pb). For the reduction reaction, where manganese is gaining electrons, you would use a manganese electrode (Mn).

3. Set up the cell: Place the lead electrode in the solution containing lead ions (Pb2+) and the manganese electrode in the solution containing manganese ions (Mn2+). Connect the electrodes with a wire and close the circuit.

4. Identify the substance being oxidized and reduced: In this case, lead (Pb) is being oxidized (loses electrons) as it is converted to lead ions (Pb2+). Manganese (Mn) is being reduced (gains electrons) as it is converted to manganese ions (Mn2+).

5. Identify the oxidizing and reducing agents: The oxidizing agent is the substance that accepts electrons, causing another substance to be oxidized. In this case, the oxidizing agent is Mn2+. The reducing agent is the substance that donates electrons, causing another substance to be reduced. In this case, the reducing agent is Pb(s).

So, if you say that both Mn and Pb are substances being reduced, that would be incorrect. Pb is being oxidized, while Mn is being reduced.

For Question 11:

To determine the products of the electrolysis of an aqueous solution of AuCl3(aq), you can follow these steps:

1. Identify the ions present: AuCl3(aq) is an aqueous solution of gold chloride. In this case, the AuCl3 dissociates into Au3+ (gold ion) and Cl- (chloride ion) in water.

2. Apply the electrolysis process: During electrolysis, positive ions migrate to the negative electrode (cathode), where they are reduced, and negative ions migrate to the positive electrode (anode), where they are oxidized.

3. Determine the reactions: At the cathode (negative electrode), the Au3+ ions will be reduced, gaining electrons to form metallic gold (Au(s)). The reduction reaction can be written as Au3+(aq) + 3e- → Au(s).

At the anode (positive electrode), chloride ions (Cl-) will be oxidized, losing electrons to form chlorine gas (Cl2(g)). The oxidation reaction can be written as 2Cl-(aq) → Cl2(g) + 2e-.

Therefore, the products of the electrolysis of the aqueous solution of AuCl3(aq) are metallic gold (Au(s)) formed at the cathode and chlorine gas (Cl2(g)) formed at the anode.