im having trouble understanding this hw question,
okay here's the question:
A 0.326 8-g unknown containing Pb(CH3CHOHCO2)2 (lead lactate,
FM 385.3) plus inert material was electrolyzed to produce
0.111 1 g of PbO2 (FM 239.2). Was the PbO2 deposited at the anode or
at the cathode?
The answers states that PbO2 is formed at the anode and is an oxidation. But when i write out the eq here:
Pb(lactate)2 --> PbO2
Pb2+ + 2e- --> Pb(s)
This is a reduction reaction...
You are both right and wrong. The second equation is wrong. The first one is right.
Pb^2+ ==> Pb02 + 2e.
The lead ion from lead lactate is +2 and it is going to PbO2 where Pb is +4. That is a loss of electrons which is oxidiation. And the PbO2 will deposit at the anode because oxidation occurs at the anode.
Ah, I see why this is confusing. It seems like chemistry is playing tricks on you! Don't worry, let me explain.
In this case, when you write out the equation: Pb(lactate)2 --> PbO2, you're right, it does look like a reduction reaction. However, this equation is not complete. It's missing an important step, which is the actual electrolysis process.
During electrolysis, oxidation occurs at the anode, while reduction occurs at the cathode. The anode is the positive electrode, where the oxidation reaction takes place, and the cathode is the negative electrode, where the reduction reaction occurs.
So, in the given scenario, although the equation you wrote might look like a reduction reaction, during the actual electrolysis process, the Pb(lactate)2 at the anode undergoes oxidation and forms PbO2. Hence, PbO2 is deposited at the anode.
Remember, sometimes chemistry likes to throw a curveball. It's all just part of the fun!
To understand why PbO2 is deposited at the anode, let's break down the electrolysis process step by step.
During electrolysis, an electric current passes through a conducting solution (electrolyte) using two electrodes: the anode and the cathode. The anode is the positively charged electrode, and the cathode is the negatively charged electrode.
In this case, the 0.3268-g unknown containing Pb(CH3CHOHCO2)2 (lead lactate, FM 385.3) is being electrolyzed. The final product of the electrolysis is 0.1111 g of PbO2 (FM 239.2).
The electrochemical reaction occurring at the anode involves the oxidation of Pb(CH3CHOHCO2)2 to form PbO2. The equation can be written as follows:
Pb(CH3CHOHCO2)2 → PbO2 + other products
This oxidation reaction is happening at the anode, so the PbO2 is deposited there. Electrolysis essentially causes the dissociation of the lead lactate compound, where the Pb(II) ions migrate towards the cathode and the lactate ions move towards the anode.
As you correctly mentioned, the reduction reaction occurring at the cathode involves the deposition of Pb metal from Pb(II) ions. The equation for this reduction reaction is:
Pb2+ + 2e- → Pb(s)
It is important to note that oxidation always occurs at the anode, and reduction always occurs at the cathode. So, in this particular case, PbO2 is formed at the anode through the oxidation reaction, and Pb metal is deposited at the cathode through the reduction reaction.
To summarize, during the electrolysis of the unknown solution containing Pb(CH3CHOHCO2)2, PbO2 is formed at the anode through oxidation, while Pb metal is deposited at the cathode through reduction.
To understand whether PbO2 is deposited at the anode or cathode, we need to analyze the oxidation and reduction reactions occurring during electrolysis.
In the given question, the unknown substance Pb(CH3CHOHCO2)2 is being electrolyzed. Electrolysis is a process where an electric current is passed through an electrolyte, causing it to undergo a chemical reaction. During this process, oxidation occurs at the anode (positive electrode), and reduction occurs at the cathode (negative electrode).
The balanced half-reaction equation for the oxidation at the anode is:
Pb(lactate)2 --> PbO2 + 2e-
On the other hand, the balanced half-reaction equation for the reduction at the cathode is:
Pb2+ + 2e- --> Pb
In the given question, it states that 0.1111 g of PbO2 (with a formula mass of 239.2 g/mol) is formed. The formula mass of Pb(CH3CHOHCO2)2 (lead lactate) is 385.3 g/mol.
To determine the mass of PbO2 formed, we can use stoichiometry. By comparing the stoichiometric coefficients, we can see that 1 mole of PbO2 is produced for every mole of Pb(lactate)2 consumed.
Let's calculate the amount of Pb(lactate)2 consumed:
0.1111 g PbO2 * (1 mol PbO2 / 239.2 g PbO2) * (1 mol Pb(lactate)2 / 1 mol PbO2) * (385.3 g Pb(lactate)2 / 1 mol Pb(lactate)2)
After performing the calculations, we find that the mass of Pb(lactate)2 consumed is approximately 0.1613 g.
Since the mass consumed at the anode (0.1613 g) is greater than the mass formed at the cathode (0.1111 g), it indicates that more substance was lost at the anode.
Therefore, PbO2 must have been deposited at the anode, suggesting that the oxidation reaction takes place there. This confirms the answer given in the solution.
Hence, based on the calculations, we can conclude that PbO2 is formed at the anode.