Electrolytic Cell: Copper Nitrate, Copper & Lead Rods

[decamij]

let's say we have an electrolytic cell with a copper (II) nitrate solution. a copper rod is the anode and a lead rod is the cathode.

Are these the reducing and oxidizing agents?
RA: NO3-
Pb(s)
Cu(s)

OA: Cu2+

Where would water go? and what are the strong reducing and oxidizing agents?

 

[GCT]

Pertaining to this equation there is only one reducing agent and one oxidizing agent, oneo f them will be in solid form while the other in anionic form. You need to compare the EMF values of each corresponding equation.

 

[decamij]

When i listed all of the RAs and OAs, i only listed them so that i can determine the strong OA and RA. By listing them, i could compare the values from a REDOX table, and see which is actually the RA and OA. So is my list correct (i.e. Would Pb and Cu metals be RAs)?

 

[Gokul43201]

Cu is definitely an RA and Cu^2+ is an OA. Pb can not be an RA because it is made the cathode.

Ask yourself what will be the result of running this cell. What will happen ? That will automatically tell you what gets reduced and what, oxidized ?

 

[decamij]

So the SRA and SOA can only be present in the solution, not at the cathodes?

 

[Gokul43201]

No, Cu from the anode is the reducing agent.

 

[GCT]

Cu is definitely an RA and Cu^2+ is an OA. Pb can not be an RA because it is made the cathode.

Ask yourself what will be the result of running this cell. What will happen ? That will automatically tell you what gets reduced and what, oxidized ?

I don't believe that the same element can be the oxidizing and the reducing agent. If copper (s) is the reducing agent, that is it is oxidized, then it will probably be oxidized by the oxidizing agent, Pb cation, which is reduced by the reducing agent.

The voltage generated will be the result of differences between the copper on one side and the Pb on the other.

 

[Gokul43201]

I don't believe that the same element can be the oxidizing and the reducing agent. If copper (s) is the reducing agent, that is it is oxidized, then it will probably be oxidized by the oxidizing agent, Pb cation, which is reduced by the reducing agent.

The voltage generated will be the result of differences between the copper on one side and the Pb on the other.

I'm not sure I follow. Where does $Pb^{2+}$ come from ? Pb makes up the cathode where only reduction can occur.

This looks more than anything, like an electrodeposition cell, where you are copper plating the lead cathode.

At the anode, $$Cu \rightarrow Cu^{2+} + 2e^-~ : oxidation$$

At the cathode, $$Cu^{2+} + 2e^- \rightarrow Cu~ : reduction$$

Perhaps I'm just wrong, it's been several years since I studied electrochemistry. What do you think the reactions at the electrodes are ?

 

[GCT]

I'm not sure I follow. Where does come from ? Pb makes up the cathode where only reduction can occur.

This looks more than anything, like an electrodeposition cell, where you are copper plating the lead cathode.

At the anode,

At the cathode,

Perhaps I'm just wrong, it's been several years since I studied electrochemistry. What do you think the reactions at the electrodes are ?

I'm not completely sure about my answer either. However, note that each of the Pb or Cu is partly soluble in the solution and thus there will be corresponding ions present. Also note that Pb is on one side of the cell while Cu on the other. For instance if copper oxidized (I'll have to check the charts to make sure of this), this means that it's electrons will flow through to reduce the Pb2+ which exists in the water, so that we see an increasing mass of Pb(s) over time, in the meanwhile voltage is generated due to the free energy of this process. I think from this you should be able to understand where I'm coming from.

 

[Gokul43201]

But the mass of Pb can not exceed the original mass of the cathode. What happens after the tiny number of Pb2+ ions in solution get reduced and redeposited onto the cathode ?

 

[GCT]

Well note that the free energy of the reaction directly relates to the relative amounts of Cu2+ that is formed by oxidation and the amount of Pb2+ that is reduced to Pb(s); that is this will probably be the limiting rate in a sense, also it will probably take a very short amount of time to reach equilbrium (pertaining to the free energy of this reaction) and for the Voltage generated to return to zero. Note that the standard reduction potentials that we use pertain to equimolar ratios of the two compounds, or in this case, probably pertains to the ratio of ions. Hope this makes sense.

 

[Gokul43201]

Well note that the free energy of the reaction directly relates to the relative amounts of Cu2+ that is formed by oxidation and the amount of Pb2+ that is reduced to Pb(s)); that is this will probably be the limiting rate in a sense, also it will probably take a very short amount of time to reach equilbrium (pertaining to the free energy of this reaction) and for the Voltage generated to return to zero.

Wait a minute. This is not a galvanic cell, it is an electrolytic cell. You are applying an external voltage across the electrodes, not requiring the spontaneous reaction to generate a voltage. I believe this may be the cause of our disagreement.

If it were a galvanic cell, I'd agree with you. Without a source of Pb2+ ions, the voltage will rapidly vanish.

 

[Gokul43201]

This, however, is a simple electroplating/electrodeposition cell. You have a cathode of some metal, an anode of fresh copper and an electrolyte containing copper ions (usually CuSo4; but in this case nitrate). At the anode the Cu gets oxidized to Cu2+, and at the cathode the Cu2+ gets reduced to copper, resulting in a copper plating over the cathode.

 

[GCT]

You're right regarding my mistake of this as an galvanic cell. I'm still a bit uncertain about your result, seems to depend a bit on the arrangment of the cell and solution used.

I'll get back on this tommorrow.

 

[buknoy]

Deposition Potential

Hi! I'm plating copper on a copper cathode and I was wondering how I can estimate the deposition potential, especially the overpotentials. Thanks

-buknoy