Here 's a cell you 'll really like The cell solution was made by mixing 25 0 mL of 4 00 mM KCN 25 0 mL of 4 00 mM KCu(CN) 2 25 0 mL of 0 400 M acid, HA, with pK a 9 50 25 0 mL of KOH solution The measured voltage was 0 440 V Calculate the molarity of the KOH solution Assume that essentially all copper(I) is Cu(CN) 2 A little HCN comes from the reaction of KCN with HA Neglect the small amount of HA consumed by reaction with HCN For the right half cell, the reaction is Cu 2 e Cu(s) 2CN Suggested procedure From E, find CN From CN , find pH From pH, figure out how much OH had been added Ag(s) AgCl(s) KCI(aq, saturated) cell solution Cu(s)

Question: Here's a cell you'll really like: The cell solution was made by mixing 25.0 mL of 4.00 mM KCN 25.0 mL of 4.00 mM KCu(CN)

Here's a cell you'll really like:

The cell solution was made by mixing

25.0 mL of 4.00 mM KCN

25.0 mL of 4.00 mM KCu(CN)₂

25.0 mL of 0.400 M acid, HA, with pK_a = 9.50

25.0 mL of KOH solution

The measured voltage was -0.440 V. Calculate the molarity of the KOH solution. Assume that essentially all copper(I) is Cu(CN)^-₂ . A little HCN comes from the reaction of KCN with HA. Neglect the small amount of HA consumed by reaction with HCN. For the right half-cell, the reaction is Cu^-₂+e^- ⇌ Cu(s) + 2CN^-. Suggested procedure: From E, find [CN]. From [CN], find pH. From pH, figure out how much OH had been added.

Ag(s)| AgCl(s)| KCI(aq, saturated) || cell solution | Cu(s)

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