If you add equation 1 to the reverse of equation 2, you will get the equation you want.
I believe you can use nFEo = RT*ln k to calculate Eo for the half cell as you have it written, then reverse it and add to Eo for equation 1 to obtain 2E for the half cell you want. Check my thinking.
Calculate the standard reduction potential for the following reaction at 25 degrees Celsius
Ag(NH3)2^+(aq) +e^- ==> Ag(s) + 2 NH3(aq) given the following thermodynamic information
Ag^+(aq) + e^- ==>Ag(s) E naught = 0.799V
Ag^+(aq) + 2NH3(aq) ==> Ag(NH3)2^+(aq) Kf = 1.7 x 10^7
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