Galvanic cell, winter 2014 final Q3

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Aschreck 2G
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Joined: Fri Sep 25, 2015 3:00 am

Galvanic cell, winter 2014 final Q3

Postby Aschreck 2G » Sat Mar 12, 2016 12:18 am

This question asks you to build a galvanic cell based on the following list of substances
Zn(s). Cu(s)
Fe(s). Pt(s)
Zn(NO3)2. Cu(NO3)2
Fe(NO3)2. Fe(NO3)3

I don't understand how you select the anode/cathode? Is it based on the highest potential difference? The question says platinum must be used as the anode, why is this?

Chem_Mod
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Re: Galvanic cell, winter 2014 final Q3

Postby Chem_Mod » Sat Mar 12, 2016 7:34 am

Please use search function to check if your question has been answered previously. This question has been addressed at least once already.

Platinum is not the anode. It is the cathode.

The answer is that a higher potential is achieved when the half reaction Fe3++e--->Fe2+ is used at the cathode. Thus, an inert Pt electrode is used simply as a surface to transfer electrons and drive this reduction. We supply the necessary ions from the aqueous solutions available. We are trying to achieve the highest cell potential.

Rebecca Choi 2H
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Joined: Fri Sep 25, 2015 3:00 am

Re: Galvanic cell, winter 2014 final Q3

Postby Rebecca Choi 2H » Sat Mar 12, 2016 5:53 pm

It makes sense that you choose Fe(NO3)3 because Fe3+ going to Fe 2+ gives the highest cell potential but then why does the answer key also circle Fe(NO3)2?


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