lectrolysis: Given the oxidation and reduction equations involved in the KI electrolysis: Reduction reaction: 2 H2O(l) + 2e- → H2(g) + 2 OH-(aq) Oxidation reaction: 2 I-(aq) →I2 (g) + 2 e- Calculate the average current that flowed through the electrolytic cell from the following information: Initial pH =7, Final pH = 12, Volume of KI solution = 30mL, Time of electrolysis is= 15min.
lectrolysis: Given the oxidation and reduction equations involved in the KI electrolysis: Reduction reaction: 2 H2O(l) + 2e- → H2(g) + 2 OH-(aq) Oxidation reaction: 2 I-(aq) →I2 (g) + 2 e- Calculate the average current that flowed through the electrolytic cell from the following information: Initial pH =7, Final pH = 12, Volume of KI solution = 30mL, Time of electrolysis is= 15min.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Given the oxidation and reduction equations involved in the KI electrolysis:
Reduction reaction:
2 H2O(l) + 2e- → H2(g) + 2 OH-(aq)
Oxidation reaction:
2 I-(aq) →I2 (g) + 2 e-
Calculate the average current that flowed through the electrolytic cell from the following information:
Initial pH =7, Final pH = 12, Volume of KI solution = 30mL,
Time of electrolysis is= 15min.

Transcribed Image Text:Electrolyti. Cell ( Non- Spontaneous)
Calculato the Average Curnont in Amps.
Amp= Coulomb/soc
f= 96,500 Coulommbs/mob
Data :
( P H=7
Final PH= 10.5
e
Time = 1omin
Vol KI solution 30.0ml
Hzot +Of
dathoda -GER
A node:LEo
Oven sll:
2I + 2H,0– I,+ Hz + OH"
Calcu lations:
-6
PoH - 3.5
[ONJ'= 3.16x10m
JOH
-6
9.5x1D mob e (96 500 Coul
mol
•92 caul.
bosm o boo Soc
99 Coulemb
boo see
.0015 Coul./soc
lomin
.3
|.5 xID A mps
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