Balancing for E

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Patrick Wilson 2B
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Joined: Wed Sep 21, 2016 3:00 pm

Balancing for E

Postby Patrick Wilson 2B » Thu Feb 02, 2017 10:43 pm

If E is intensive and we don't multiply the value of it by a coefficient when balancing a reaction, why do we need to balance it at all? Also in an example from lecture Copper becomes Copper(2) and its E=-0.34 . However, when we solve for Ecell we use the positive value of 0.34 instead even though the redox reaction says Copper is losing electrons. Shouldn't it be Ecell=0.77 - (-0.34)=1.11 ? Thanks in advance.

Edward_Lee_3C
Posts: 36
Joined: Wed Sep 21, 2016 2:58 pm

Re: Balancing for E

Postby Edward_Lee_3C » Fri Feb 03, 2017 12:50 am

If you watch the bruincast for wednesday he explains why "we use the positive value of 0.34 instead even though the redox reaction says Copper is losing electrons. Shouldn't it be Ecell=0.77 - (-0.34)=1.11 ? ". We also balance it to cancel out the electrons and balancing our equations will help when we calculate G(standard) = -nFE(standard)

Vincent Tse 2B
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Joined: Fri Jul 22, 2016 3:00 am
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Re: Balancing for E

Postby Vincent Tse 2B » Fri Feb 03, 2017 11:40 am

Typically, balancing a redox reaction ensures that you are accounting for the correct amount of electrons being transferred from one species to another.

It is this movement of electrons that generates the energy used for work in terms of voltage (in this case, emf or E). When calculating this available energy, we consider Gibb's Free Energy and the formula deltaG(std) = -nFE(std) (where n = the mole of electrons, F = the charge per mol of electron using the units of Faraday, and E is the standard reduction potential). By balancing the redox reaction, that is how we will determine the correct release of energy (or absorption) in the system.


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