You are given the following half-cell reactions: Pd 2+ (aq) + 2e Pd(s)E o =
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You are given the following half-cell reactions:
Pd2+ (aq) + 2e– → Pd(s)………………………Eo= = 0.83 V
PdCl2–4 (aq) + 2e– → Pd (s) + 4Cl– (aq)……Eo = 0.64 V
a. Calculate the equilibrium constant for the reaction
Pd2+ (aq) + 4Cl– (aq) ⇋ PdCl2–4(aq)
b. Calculate ΔGo for this reaction.
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a b G o nFE o 2 96 485 C mol 1 019 V 367 kJ mo...View the full answer
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