Writing Balanced Equations for Galvanic Cells (14.11 part d sixth edition)

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KatelinTanjuaquio 1L
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Writing Balanced Equations for Galvanic Cells (14.11 part d sixth edition)

Postby KatelinTanjuaquio 1L » Mon Feb 25, 2019 9:25 pm

Question 14.11 of the sixth edition, part d states:

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells. Pt(s) | O2(g) | H+ (aq) || OH-(aq) | O2(g) | Pt(s).

I understand how the anode reaction is derived, but how is the cathode reaction found? Whenever I try, my answer does not match the one in the solutions manual.

josie 1C
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Re: Writing Balanced Equations for Galvanic Cells (14.11 part d sixth edition)

Postby josie 1C » Tue Feb 26, 2019 8:47 am

you have to make sure that you're writing the cathode as a reduction reaction, so the product should have less charge than the reactant

Kessandra Ng 1K
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Re: Writing Balanced Equations for Galvanic Cells (14.11 part d sixth edition)

Postby Kessandra Ng 1K » Tue Feb 26, 2019 9:40 am

To add onto what the user above said, what to remember is that because it's reduction, you add on electrons onto the left side of the equation.

Charles Hood Disc 1C
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Re: Writing Balanced Equations for Galvanic Cells (14.11 part d sixth edition)

Postby Charles Hood Disc 1C » Tue Feb 26, 2019 10:43 am

For the cathode, try to imagine the Oxygen as being the electrons supplied to make the OH-. When it is added to water it will increase the number of electrons and form OH-. (Really it is the electrons adding on with the oxygen)


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