H+ and e-

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Laine Gruver 3C
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Joined: Wed Sep 21, 2016 2:56 pm

H+ and e-

Postby Laine Gruver 3C » Tue Feb 14, 2017 7:15 pm

When balancing redox half reactions and using e- and H+ to even out the charges on both sides of the equation, why do the H+ and e- not cancel each other out if they are on the same side of the equation?

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Joined: Wed Sep 21, 2016 2:58 pm

Re: H+ and e-

Postby Hannah_El-Sabrout_2K » Tue Feb 14, 2017 9:08 pm

They do because one is +1 and the other is -1, so they add up to zero. Is there a specific question that you're referencing?

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Joined: Wed Sep 21, 2016 2:57 pm

Re: H+ and e-

Postby samuelkharpatin2b » Fri Feb 17, 2017 3:08 pm

The charges which come from the H+ and e- do cancel each other out, you just have to add an appropriate amount of each so that both sides of the reaction are equal in charge and in the amount of atoms.

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Re: H+ and e-

Postby Tycho_Meimban_2B » Fri Feb 17, 2017 4:42 pm

H+ is actually used to balance out the "H" molecules in the equation, but e- electrons are needed to balance the charge, especially after (H20)aq (for aqueous) or (OH-)aq (for basic) are used to balance the Oxygen molecules. The electrons you use are needed to balance out the charge of the products and the reactants.

The H+ is needed to balance the atoms in equation, while the electrons are used to balance the charges.

Hope this helps!

Vivian Tan 3F
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Joined: Fri Sep 25, 2015 3:00 am

Re: H+ and e-

Postby Vivian Tan 3F » Sun Feb 19, 2017 2:42 am

The H+ protons are to balance the H atoms in the equation and the electrons are used to balance the charge.

For example, to balance this equation:
2e- + SeO42- --> SeO32- + H2O

You will need to add 2 H+ ions to the left to balance out the 2 H molecules in H2O.

This will form:
2 H+ + 2e- + SeO42- --> SeO32- + H2O

Adjust the charges accordingly -- the left side of the equation now has a 2- charge ( 2+(-2)+(-2) ) and the right side also has a 2- charge. This is balanced.

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