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A 0.200 M KOH solutions is electrolyzed for 1.5 h using a current of 8.00 A. How many moles of O₂were produced at the anode?

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0.48

0.224

0.112

2.24X10^-2

answer is C.

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Detailed Solution

$m = \frac{E}{96500} X i XT$

$m = \frac{8}{96500} X 8 X 1.5 X 60 X 60$

$m = 3.5$

$no . of mole = \frac{3.5}{32} = 0.1119$

$= 0.112$

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