6h Edition; 14.23

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6h Edition; 14.23

Postby JeremyPark14B » Wed Feb 27, 2019 11:43 pm

In part a, the half reaction within the main reaction is 6Hg -> 3Hg2 2+ + 6e-

In the reduction potential E values, Hg2 2+ + 2e- -> 2Hg E = +.79.

Why are we to use the +.79 instead of flipping the sign to -.79 since the oxidation version of this reaction is done in the actual reaction?

Vy Lu 2B
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Re: 6h Edition; 14.23

Postby Vy Lu 2B » Sat Mar 02, 2019 5:04 pm

To calculate the cell potential of the reaction, there are two methods in doing so; one is not switching any signs on any of the standard reduction potentials and simply subtracting them in this manner (cathode - anode) or the other method is flipping the anode potential's sign and then adding the two potentials together. Overall, you can utilize any of the two methods to obtain your cell potential.

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Re: 6h Edition; 14.23

Postby AdityaGuru1H » Sat Mar 02, 2019 6:10 pm

to make sure that you don't confuse your signs it may be best to figure out which method you like the best and stick with that otherwise you may get wrong answers. A good way to check is to see if you get a positive answer as most of the cells should end up having a positive potential

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