## Problem 14.15

104822659
Posts: 23
Joined: Wed Sep 21, 2016 2:57 pm

### Problem 14.15

Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions:
(a) $AgBr(s)\rightleftharpoons Ag^{+} (aq)+Br^{-}(aq)$, a solubility equilibrium
(b)$H^{+}(aq)+OH^{-}(aq)\rightarrow H_{2}O(l)$, the Brønsted neutralization reaction
(c) $Cd(s)+2Ni(OH)_{3}(s)\rightarrow Cd(OH)_{2}(s)+2Ni(OH)_{2}(s)$, 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?

lilyjustine_1G
Posts: 10
Joined: Fri Jun 17, 2016 11:28 am

### Re: Problem 14.15

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
Posts: 10
Joined: Wed Sep 21, 2016 2:55 pm

### Re: Problem 14.15

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
Posts: 15
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Problem 14.15

@lilyjustine_1G

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.

Esther_Choe_1K
Posts: 11
Joined: Fri Jun 17, 2016 11:28 am

### Re: Problem 14.15

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!