oxidation reduction question?

[thatdumbstudent]

Hi guys, from the attached information, I have to find the oxidising and reducing agent then write the half equations for anode and cathode. I am confused...

I believe that the oxidant is Cu2+ and reductant is Cr2O72- , is that right? I just need the clarification then I can write the half eqns by myself, thanks!

[thatdumbstudent]

Hi guys, from the attached information, I have to find the oxidising and reducing agent then write the half equations for anode and cathode. I am confused...

I believe that the oxidant is Cu2+ and reductant is Cr2O72- , is that right? I just need the clarification then I can write the half eqns by myself, thanks!

wait or maybe I've swapped them... because the voltage can't be negative, the copper would be oxidized?

[Geoo]

Hi guys, from the attached information, I have to find the oxidising and reducing agent then write the half equations for anode and cathode. I am confused...

I believe that the oxidant is Cu2+ and reductant is Cr2O72- , is that right? I just need the clarification then I can write the half eqns by myself, thanks!

So, both E nort values are positive and are out of order, so you swap the equations around so that it is 1.33 V insert dichromate equation here, and then 0.34 V insert copper equation here.

So now that you have the order, you need to find what will spontaneously react, so that you can work out the oxidant and reductant.

In this case the reductant is Cu (s), as it is oxidises and the oxidant is the Dichromate ion (Cr2O72-) as it undergoes reduction itself.

So you were right the first time. For the record, voltage can be negative in certain cases where it has more negative polarity than the circuit. Other wise the total potential difference of some cells wouldn't make sense of voltage couldn't be negative. Hope this helps.

[Ionic Doc]

Hi guys, from the attached information, I have to find the oxidising and reducing agent then write the half equations for anode and cathode. I am confused...

I believe that the oxidant is Cu2+ and reductant is Cr2O72- , is that right? I just need the clarification then I can write the half eqns by myself, thanks!

IDK if I'm correct, but the way I think about it is that the equation with the higher E nort value undergoes reduction while the equation with the lower E nort value undergoes oxidation in a galvanic cell. You also have to keep in mind that the strongest oxidising agent (top left) reacts with the strongest reducing agent (bottom right).

So, in this case, the dichromate equation undergoes reduction (which also means the reaction will occur at the cathode). Cr₂O₇^2-will also be the oxidising agent. 

Then, this means the copper equation undergoes oxidation (which also means the reaction will occur at the anode). Cu will also be the reducing agent. 

And you can use the electrochemical series to determine the half equations for these reactions. Keep in mind that you have to flip the half equation for the copper reaction as it undergoes oxidation not reduction (which is what the elctrochemical series reaction are written as)

[thatdumbstudent]

So, both E nort values are positive and are out of order, so you swap the equations around so that it is 1.33 V insert dichromate equation here, and then 0.34 V insert copper equation here.

So now that you have the order, you need to find what will spontaneously react, so that you can work out the oxidant and reductant.

In this case the reductant is Cu (s), as it is oxidises and the oxidant is the Dichromate ion (Cr2O72-) as it undergoes reduction itself.

So you were right the first time. For the record, voltage can be negative in certain cases where it has more negative polarity than the circuit. Other wise the total potential difference of some cells wouldn't make sense of voltage couldn't be negative. Hope this helps.

This helps! Thank you! :-)

[thatdumbstudent]

IDK if I'm correct, but the way I think about it is that the equation with the higher E nort value undergoes reduction while the equation with the lower E nort value undergoes oxidation in a galvanic cell. You also have to keep in mind that the strongest oxidising agent (top left) reacts with the strongest reducing agent (bottom right).

So, in this case, the dichromate equation undergoes reduction (which also means the reaction will occur at the cathode). Cr₂O₇^2-will also be the oxidising agent. 

Then, this means the copper equation undergoes oxidation (which also means the reaction will occur at the anode). Cu will also be the reducing agent. 

And you can use the electrochemical series to determine the half equations for these reactions. Keep in mind that you have to flip the half equation for the copper reaction as it undergoes oxidation not reduction (which is what the elctrochemical series reaction are written as)

Ahh okay this makes much more sense now, thank you!!