8/19/2019 Answer Key Homework 7 (1)

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Homework 7

1 O2 consumed by a 70-kg human is about 16mol per day. This O2 oidi!es "ood and is reduced to H2O #ia the "ollowing reaction$O2 % &H

% % &e '2H2O To what current (assume the current is constant) does this

respiration rate correspond* Answer: According to O2 + 4H

+  + 4e ⇋ 2H2O, the mole ratio between O2 

and electrons is 1:4, therefor, 16 mole O2 is equialent to 16!4

mole electrons" #he charge of 1mole electron is $6%&& ', so total

charges of 16!4 mole electrons are 16!4!$6%&& '" since current equals charge(time in second, and one da) has 24!6&!6&

seconds, this gies a current of 16!4!$6%&& '( *24!6&!6& s"

2 How many hours are re+uired "or 0.100 mole o" electrons ,ow

through a circuit i" the current is 1.00

 Answer:#he charge of 1mole electron is $6%&& ', so total charges of

&"1&& mole electrons are &"1&&!$6%&& '" since current equals

charge(time in second, this gies the time in seconds of

&"1&&!$6%&& '( *1"&& A"

The chlor-alkali process/ in which seawater is electroly!ed to

produce l2 and aOH/ is the second most important commercial

electrolysis/ behind production o" luminum. The "ollowing is the

reaction occurred on the positi#e electrode$2l- -2e  l2" 1.364105  electrons ,ow through the electrode/ how many

moles o" l2 can be produced* Answer:

8/19/2019 Answer Key Homework 7 (1)

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#he charge of 1mole electron is $6%&& ', so moles of 1.364105 

electrons is 1.364105  36500.ccording to 2l -  -2e   l2/ two moles electrons is e+ui#alent to

1mole chlorine. o you should be able to 8gure out how to get

moles o" chlorine.

& n the coulometric titration o" 9&% in the presence o" ecess e%/it was "ound to re+uire 652 seconds to reach the e+ui#alencepoint using a constant current o" 100.0 m. How many moles o"9&% were present in the solution* (Hints$ split the "ollowingreaction into oidation and reduction hal"/ then/ 8nd therelationship between moles o" electron trans"erred and moles o"

9&%

 Answer:n this titration, 'e4+ is formed on the electrode surface ia the

following electrode reaction:'e-+ .e   'e4+

#he o/idation and reduction half reaction of the following reaction

is:2'e4+  +2e   2'e-+  *reduction04+  .2e +2H2O   0O2 

2+  + 4H+  *o/idation

'harge goes through the electrode is 6%2!&"1&&&& '" this

amount of charge is equialent to 6%2!&"1&&&&($6%&& mole"

ince 2 mole electrons is equialent to 1 mole 04+ *see the

o/idation half, so the mole of 04+ 6%2!&"1&&&&($6%&&(2