9.15 cell diagrams  [ENDORSED]

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Jessica Lutz 2E
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9.15 cell diagrams

Postby Jessica Lutz 2E » Mon Feb 19, 2018 10:35 am

I am confused with multiple parts of this problem in regards to how to write out the cell diagram:
Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions:
(a) AgBr(s) <--> Ag+ (aq) + Br- (aq), a solubility equilibrium
(b) H+ (aq) + OH- (aq) --> H2O(l), the Brønsted neutralization reaction
(c) Cd(s) + 2 Ni(OH)3(s) --> Cd(OH)2(s) + 2Ni(OH)2(s), the reaction in the nickel–cadmium cell

For part a, does the order of the compounds on the anode side of the diagram matter. I thought it was supposed to go from reactants to products, but the solutions manual has the order: Ag(s)|AgBr(s)|Br-(aq). Does that have something to do with the fact that Ag and AgBr are both solid while Br- is aqueous?
For part b, why do you leave water out of the cell diagram?
For part c, how did you know to include a salt in it when we never did in cell diagrams previous to this one? Also, did we just use Ni(s) as the conducting electrode instead of Pt(s) because Ni is already involved in the reaction?

Help on any of these questions would be much appreciated!

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Re: 9.15 cell diagrams  [ENDORSED]

Postby Chem_Mod » Wed Feb 21, 2018 2:54 pm

A) You are correct. First separate by phase, then separate by reactants and products. For cell diagram, within each half part, electrode and solids are written far away from the salt bridge, followed by liquid, gas and aqueous solution. Within each phase, orders are not strictly defined, but people tend to put in the order of reactant to product.

B) Water was likely excluded because the other reactants were labeled as aqueous, meaning you can assume water was involved in the reaction.

C) Yes, nickel was probably used because it was already involved in the reaction and therefore you will not have any side reactions with platinum. However, Platinum would work as well. See this question for an explanation for KOH: viewtopic.php?f=140&t=11305&p=86457&sid=44003ca558573b09df76acd11c2cc82f#p86457

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