galvanic cell question

[gangazi]

i was trying to do 1.82V-0.53V but not in the answer choice... can anyone explain this for me thank you!

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[Ashish]

For any cell, one half of reaction must be oxidation and the other half must of reduction (An Ox and Red Cat). Here, Non metal Iodine is the reduction half rxn. So the Cobalt half must be oxidation half (makes sense because Cobalt is a transition metal and Iodine is a non metal). However, the reduction half is given for Cobalt so we need reverse it to make oxidation reaction which reverses the E cell to -1.82 V. Then its simple plug and chug:

E cell= E ox + E red =-1.82+0.53 = -1.29 V

 

[rdub]

@Ashish Isn't the Ecell for a galvanic cell always positive though? Thats what I have in my notes at least...

 

[Ashish]

@Ashish Isn't the Ecell for a galvanic cell always positive though? Thats what I have in my notes at least...

Yeah thats true. To make that happen I guess Iodine is the anode? But that leads to answer that is not even one of the options? I am stumped lol We need @orgoman22 to the rescue. Like always.

 

[super frank]

Probably a faulty question... Most positive value written in terms of reduction is the reduction potential. Then flip the less positive, which will then equal the oxidation potential. E = red + ox. Supported by the fact that galv/volt cells are also always spontaneous (Ecell > 0).

What software is it from?

 

[DAT Destroyer]

View attachment 204334

i was trying to do 1.82V-0.53V but not in the answer choice... can anyone explain this for me thank you!

In a galvanic cell the overall cell voltage must be positive. You are given 2 half reactions that are both reductions, thus one of them must be reversed. Keeping in mind that the overall answer MUST be positive. Keep the first equation and reverse the 2nd equation. 1.82 - .53 will do the job. I suggest you consult a general chemistry textbook written by a PhD chemist to reinforce the concept.

Hope this helps..

Dr. Romano

 

[Minions2015]

In a galvanic cell the overall cell voltage must be positive. You are given 2 half reactions that are both reductions, thus one of them must be reversed. Keeping in mind that the overall answer MUST be positive. Keep the first equation and reverse the 2nd equation. 1.82 - .53 will do the job. I suggest you consult a general chemistry textbook written by a PhD chemist to reinforce the concept.

Hope this helps..

Dr. Romano

Hey, so does the opposite apply to electrolytic cells?

 

[super frank]

Hey, so does the opposite apply to electrolytic cells?

In electrolysis the cation will be reduced and the anion will be oxidized. The potentials given will reinforce the fact that the reaction is non-spontaneous (<0).

 

[DAT Destroyer]

Hey, so does the opposite apply to electrolytic cells?

That is correct, keep hammering away you are on the right track.

Dr. Romano