following argentometric titration reaction was monitored using an electrochemica
ID: 103928 • Letter: F
Question
following argentometric titration reaction was monitored using an electrochemical cell. Ags) wire with a Agl (s) coating as the cathode and an SCE as the anode. Titration reaction: Ag+(a) + raq) Agl (s) Monitoring cell: SCE I| I' (aq) | Agl(s), Ag (s) (a) (4 pts) Derive a "split" Nernst equation that describes the monitoring cell for this titration. Plug in the Ksp for Agl-8.3 x 10- Eocat-0.152 V, Eoan = 0.241 numbers for any variables that are known. Hint: To get started, write out the cathodic half reaction (b) (6 pts) If 50.00 mL of 0.1500 M KI was reacted with 20.00 ml of 0.2800 M AgNO, what would cell potential be? (c) (6 pts) If 50.00 mL of 0.1500 M KI was reacted with 40 mL of 0.2800 M AgNO,, what would the cell potential be?Explanation / Answer
a)
AgI(s) + e Ag(s) + I
E0cat = 0.152 V
Ecat = E0cat – RT / nF ln([I-])
E0an of SCE = 0.241
At 25 C,
Ecell = Ecat + Ean
= 0.241 – 0.152 – 0.05912 / 1 * log([I-])
= 0.089 - 0.05912 * log([I-])
b)
KI + AgNO3 --> KNO3 + AgI(s)
Moles if I- left in solution = 50 * 0.15 – 20 * 0.28
= 1.9 mMoles
Total volume of solution = 50 + 20 = 70 mL
[I-] = 1.9 / 70 = 0.027 M
Ecell = 0.089 - 0.05912 * log([I-])
= 0.089 – 0.05912 * log(0.027)
= 0.18 V
c)
Moles if Ag+ left in solution = 40 * 0.28 - 50 * 0.15
= 3.7 mMoles
Total volume of solution = 50 + 40 = 90 mL
[Ag+] = 3.7 / 90 = 0.041 M
Ksp = [Ag+][I-] = 8.3 x 10-17
[I-] = 8.3 x 10-17/ 0.041
= 2.02 x 10-15 M
Ecell = 0.089 - 0.05912 * log([I-])
= 0.089 – 0.05912 * log(2.02 x 10-15)
= 0.96 V
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