The following cell has a voltage of 0.083 V: Hg(l) | Hg(NO3)2(0.001 0 M), KI(0.500 M)
Question:
The following cell has a voltage of 0.083 V:
Hg(l) | ƒ Hg(NO3)2(0.001 0 M), KI(0.500 M) || S.H.E.
From this voltage, calculate the equilibrium constant for the reaction
In 0.5 M KI, virtually all the mercury is present as HgI24-.
Transcribed Image Text:
Hg²+ + 41 = Hgl?
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