In example 13.18 in the text book, we have the reaction AgCl ----> Ag(+) + Cl(-).
therefore the half reactions are:
AgCl + e ---> Ag + Cl(-) +0.22V
Ag(+) + e ----> Ag +0.80V
we then reverse the second equation to get the original reaction.
However, why dont we also flip the sign on the E when we calculate the E for the cell?
Example 13.18 - Text book
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Re: Example 13.18 - Text book
The textbook uses the equation E°cell=E°cathode-E°anode. When using this equation, the minus sign accounts for flipping the sign of the anode's reduction potential. Notice only reduction potentials are used when using this equation.
If you flip the sign of the anode, which is the method I personally prefer, it makes more sense what's happening. In this case, you use the equation E°cell=E°oxidation+E°reduction, where the oxidation half reaction has what was previously its standard reduction potential now as an oxidation potential (the sign has been changed after flipping the reaction).
Both methods yield the same final standard cell potential.
If you flip the sign of the anode, which is the method I personally prefer, it makes more sense what's happening. In this case, you use the equation E°cell=E°oxidation+E°reduction, where the oxidation half reaction has what was previously its standard reduction potential now as an oxidation potential (the sign has been changed after flipping the reaction).
Both methods yield the same final standard cell potential.
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