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Calculating EMF of a cell

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Awuah29

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What is the standard EMF if the following half reactions are combined in a galvanic cell?

(Equation 1) Co3+( aq ) + e ---- Co2+ (aq) E= 1.82 V

(Equation 2 ) Na+ (aq)+ e ----- Na (s) E= -2.71V

Ecell = Ered+Eox We have to reverse the first equation to –1.82 and also the second equation + 2.71
so –1.82+(+2.71)= 0.89
Please correct me if I am wrong
Thanks
 

Lonely Sol

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Hey man, you are absolutely right!! If it is a galvanic cell, the EMF has to be "+" and if it is a electrolytic cell, then EMF = "-". Also free energy is opposit of emf for the cell. Here is a quick chart:

Electrolytic Galvanic
Cathode - +
Anode + -
EMF - +
delta G + -

I am pretty sure about it. If anyone thinks its wrong, please feel free to corect me!
 

poc91nc

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This is how I would solve the problem…

Flipping the sodium equation as you did gives:
Na (s) ----- Na+(aq) + e- gives E = 2.71V

Flipping the second equation gives you the following:
Co2+ ------ Co3+ (aq) + e- gives E = -1.82

Look at the format of the equations, isn’t there something wrong? How is it possible to have two oxidations occurring??? Oxidation always occurs at the anode, and reduction occurs at the cathode. You would be losing solid sodium at the anode, and obtaining solid Co at the cathode. This would all be from electron transfer. Therefore, you cannot flip the Co equation, and the E cell = 2.71 + 1.82
 

poc91nc

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You are right, the E has to be positive for a galvanic cell...but given the values there is more than one way to flip/add the numbers to get a positive value...the first configuration presented in the original post has two oxidations occuring (based on the sign values used in addition). At least that is my reasoning.
 

BenignDMD

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Since this is a galvanic cell, the EMF must be positive. The E values you gave are the values for the reduction potential of each half reaction. The cobalt half reaction has a more positive reduction potential and thus would be the reduction half reaction (thus cobalt is the oxidizing agent, or oxidant). The sodium half reaction would be reversed b/c it would be the oxidation half reaction in this case. You flip the sign as well then add both E values together, giving you 4.53V.

In an electrolytic cell the opposite would occur and the EMF would be -4.53 V.

Hope that helps. I am a Kaplan instructor and this is one of our lesson book problems.
 
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