I need help with Reduction Potentials please!!!

[Dr.9999]

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Can someone pls help me with this problem. I don't understand the answer the picked for this ques. Thanks

A voltaic cell is assembled using Zn and Ni with a salt bridge. The reduction potentials are given below. Choose the true statement?

Zn2+(aq)+2e−⟶Zn(s) Eo= −0.76 V
2Ni+(aq)+2e−⟶Ni(s) Eo= −0.28 V

A) Zn will be the cathode and Ni will be the anode
B) It is impossible to determine the cathode and anode without the concentrations of Zn+2 and Ni+2
C) Zn will be the anode and Ni will be the cathode
D) Zn+2 will be the anode and Ni+ will be the cathode
E) Zn+2 will the cathode and Zn will be the anode

 

[steppingstone]

what is the answer? i can explain it to you but I don't want to explain the wrong answer!

 

[Papasmurf1234]

Is it C?

Red Cat/ An Ox.

For Galvanic/Voltaic cells REDUCTION occurs at the CATHODE
And OXIDATION occurs at the ANODE.

Ni equation has a higher (more positive) reduction potential so it will be more likely reduced than the Zinc.

 

[ReaverOpS]

C is the right answer. This was a question from datQvault.

 

[Dr.9999]

Yes the answer is C. However I don't understand what they mean by Ni+? Is it the opposite direction so -.28 then become +.28?

 

[Papasmurf1234]

Yes the answer is C. However I don't understand what they mean by Ni+? Is it the opposite direction so -.28 then become +.28?

They're talking about the ion.

In a galvanic cell, because the Eo is higher (more positive) for the Nickel (Ni+), it will be reduced.

Easier way to learn this is like this:

If I had these data:

Zn2+(aq)+2e−⟶Zn(s) Eo= 0.76 V
2Ni+(aq)+2e−⟶Ni(s) Eo= 0.28 V

Then the Zn2+ has a more positive aka higher number value with the reduction equation because the electrons are on the left side. So therefore, comparing those two ions, Zn2+ will be reduced and 2Ni+ will be oxidized.

If I had a data like this:

Zn2+(aq)+2e−⟶Zn(s) Eo= 0.28 V
2Ni+(aq)+2e−⟶Ni(s) Eo= 0.76 V

Then the Ni+ is more positive aka higher number so it will most likely be reduced while the Zn2+ will be oxidized.

Now...if my data were like this:

Zn2+(aq)+2e−⟶Zn(s) Eo= -0.76 V
2Ni+(aq)+2e−⟶Ni(s) Eo= -0.28 V

Then the Ni+ value is more positive aka higher number so Ni+ will most likely be reduced.

Now..one more...
If my data were like this:

Zn2+(aq)⟶Zn(s)+ 2e− Eo= 0.76 V
2Ni+(aq)⟶Ni(s) +2e- Eo= 0.28 V

These are oxidation reactions because the electrons are on the right side - aka the electrons come off (Oxidation is loss). Therefore, to find the reduction potential you have to reverse the signs.

If you do that, the equations will become:

Zn2+(aq)+ 2e− ⟶Zn(s) Eo= -0.76 V
2Ni+(aq) +2e- ⟶Ni(s) Eo= -0.28 V

So...in conclusion, given the equation, whether oxidation or reduction, the more positive/higher number will be favored.

And in order to find the EMF you use this equation.

Ered+Eox = EMF



Now for the quiz....


If my equations were like this:

Zn2+(aq)+ 2e− ⟶Zn(s) Eo= -0.76 V
2Ni+(aq) +2e- ⟶Ni(s) Eo= -0.28 V

You know that Ni+ is reduced because the Eo is more positive/higher and the reaction given is a reduction equation.
And you know that you need to flip the Zn2+ equation to get the oxidation equation.

So...

Ered + Eox = EMF

-0.28 + (+0.76) = 0.48.

I literally broke down every step so hopefully this will help you get the concept down. Its very intuitive if you think about it.

 

[Papasmurf1234]

^^ wow that really helped me understand the concept. Now I need to bookmark this Page. Thanks a lot.

Sent from my Samsung Galaxy S2!

 

[Dr.9999]

Thank you!