Suppose I have a Pt disk electrode in a solution of Cu 2 plus 0 1 M KCl versus a saturated calomel electrode (ESCE 0 242 V) If the E 1 2 for Cu 2 reduction is at 0 100 V vs SCE, what would the E 1 2 be if I added 0 10 M NH 3 and the following reaction occurred Cu 2 4NH 3 Cu(NH 3 ) 4 2 K 1 1 x 1013

Question: Suppose I have a Pt disk electrode in a solution of Cu 2+ plus 0.1 M KCl versus a saturated calomel electrode (ESCE = 0.242

Suppose I have a Pt disk electrode in a solution of Cu₂₊ plus 0.1 M KCl versus a saturated calomel electrode (ESCE = 0.242 V). If the E_1/2 for Cu²⁺ reduction is at 0.100 V vs. SCE, what would the E_1/2 be if I added 0.10 M NH₃ and the following reaction occurred:

Cu²⁺ + 4NH₃----------- Cu(NH₃)₄²⁺ K = 1.1 x 1013

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