Standard reduction potentials - Boston University

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Standard reduction potentials. 3/24/2013 10:02 AM. Copyright © 2013 Dan Dill [email protected]. 1. Standard reduction potentia
Standard reduction potentials

3/24/2013 10:02 AM

Lecture 17 CH102 Spring 2013

Copyright © 2013 Dan Dill [email protected]

Standard reduction potentials Boston University Slideshow Title Goes Here

Standard reduction potentials

Standard Hydrogen Electrode (SHE) has Eo = 0 V 2 H+(1 M) + 2 e‐  H2(1 atm), Eo = 0 V All other reductions defined relative to SHE Zn2+(1 M) + 2 e‐  Zn(s), Eo = –0.763 V Cu2+(1 M) + 2 e‐  Cu(s), Eo = +0.340 V etc.

CH102 Spring 2013 Boston University

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Lecture 17 CH102 Spring 2013

Copyright © 2013 Dan Dill [email protected]

Combining reduction potentials

Copyright © 2013 Dan Dill [email protected]

Combining reduction potentials

Boston University Slideshow Title Goes Here

Boston University Slideshow Title Goes Here

Eo = Eocathode – Eoanode Why? o ΔG cell = ΔGocathode + ΔGoanode −n F Eocell = −n F Eocathode − (−n F Eoanode) … “−” since anode runs as an oxidation Eocell = Eocathode − Eoanode

 = –0.763 V 2+ ‐ o Cu (1 M) + 2 e  Cu(s), E = +0.340 V Cu2+(1 M) + Zn(s)  Cu(s) + Zn2+(1 M), Eo = ? Eo = Eocathode – Eoanode = +0.340 V – (–0.763 V) = +1.103 V So, Zn is oxidized by Cu2+ Zn2+(1 M) + 2 e‐

Lecture 17 CH102 Spring 2013

Zn(s), Eo

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Copyright © 2013 Dan Dill [email protected]

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Standard reduction potentials

Lecture 17 CH102 Spring 2013

3/24/2013 10:02 AM

Copyright © 2013 Dan Dill [email protected]

Combining reduction potentials

Copyright © 2013 Dan Dill [email protected]

Combining reduction potentials

Boston University Slideshow Title Goes Here

Boston University Slideshow Title Goes Here

When result does not contain e‐ … Cu2+(1 M) + Zn(s)  Cu(s) + Zn2+(1 M)

Fe3+(1 M) + e‐  Fe2+(1 M ), Eoa = +0.771 V,  Fe2+(1 M) + 2 e‐  Fe(s), Eob = −0.409 V,  Fe3+(1 M) + 3 e‐  Fe(s), Eo = ? ΔGo = ΔGoa + ΔGob −3 F Eo = −F Eoa + (−2 F Eob) … since both are reductions Eo = Eoa/3 + 2 Eob/3 = (+0.771 V)/3 + 2(−0.409 V)/3 = −0.016 V

Eo = Eocathode – Eoanode When result does contain e‐ … Fe3+(1 M) + 3 e‐  Fe(s),

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Copyright © 2013 Dan Dill [email protected]

Lecture 17 CH102 Spring 2013

use ΔGo !!!

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