Problem 14.15

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Problem 14.15

Postby 104822659 » Sat Feb 11, 2017 5:47 pm

Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions:
(a) , a solubility equilibrium
(b), the Brønsted neutralization reaction
(c) , the reaction in the nickel–cadmium cell

For 15A, why does the book flip AgBr(s)? Why do we have to reverse the anode? Also, how come the answer for 15C in the solution manual includes a Potassium? Where did that come from?

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Re: Problem 14.15

Postby lilyjustine_1G » Sat Feb 11, 2017 6:37 pm

In additon, for 15c, why did we only add an aqueous solution like KOH to the anode side and not the cathode side?

Ashley Shirriff 2F
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Re: Problem 14.15

Postby Ashley Shirriff 2F » Sat Feb 11, 2017 6:44 pm

Adding to this, the solution manual has Ni(s) on the cathode side, even though it was not part of the final reaction. I understand that it is used as the conducting electrode for the cathode, however, how did they know to use Ni? And how do we know when we have to add a conducting electrode?

David Sung 2H
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Re: Problem 14.15

Postby David Sung 2H » Sun Feb 12, 2017 3:32 pm


I think the KOH is added to the cell diagram because an aqueous solution consisting of OH- ions is needed to supply the reaction with OH- ions. It's not added to the cathode side because Ni(OH)3 can supply with the OH- ions to Cd(OH)2 by itself. Not completely sure but this is the only explanation I can come up with.

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Re: Problem 14.15

Postby Esther_Choe_1K » Sun Feb 12, 2017 4:47 pm

For Question 14.15 part A, I am not sure how the half reactions are obtained, particularly the oxidation half reaction. I know that bromine is being oxidized from 1- to 0 but how do we know exactly that the Ag added to the Br- is the solid Ag and not the aqueous Ag? Thank you for any help!

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