14.1 Part D

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Jessica Nunez 1I
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Joined: Thu Jul 13, 2017 3:00 am

14.1 Part D

Postby Jessica Nunez 1I » Sat Feb 17, 2018 10:54 pm

I am confused as to how the chemical equation on 14.1, Part D was balanced.

Sungyoon_Baek_1A
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Re: 14.1 Part D

Postby Sungyoon_Baek_1A » Sat Feb 17, 2018 11:08 pm

First we add the two half reactions from part b and c (remember to multiply part b by 3 so that the electrons lost in part b matches the electrons gained in part c). Everything should be balanced. We see that there are 14 H+ on the left and 6H+ on the right so we can cancel all of the H+ on the right and we are left with 8H+ on the left and that is the final answer.

MorganYun1H
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Joined: Thu Jul 13, 2017 3:00 am

Re: 14.1 Part D

Postby MorganYun1H » Sat Feb 17, 2018 11:11 pm

you have to split it up into its half reactions. but when you add up the equations, you have to make sure to multiply each half reaction to make the number of electrons on each side cancel out.

Xin He 2L
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Re: 14.1 Part D

Postby Xin He 2L » Sun Feb 18, 2018 1:23 am

To balance an equation, you split them up into oxidation and reduction half equations. Then you balance those equations out with electrons and then make sure to cancel the electrons when you add the equations back together.

Sandhya Rajkumar 1C
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Re: 14.1 Part D

Postby Sandhya Rajkumar 1C » Sun Feb 18, 2018 12:15 pm

After obtaining the half-reactions, you have to multiply the oxidation half-reaction (part b) by 3 so that both half-reactions have the same number of electrons involved. Afterwards, you combine the two equations.


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