April 20, 2024, 04:12:25 AM
Forum Rules: Read This Before Posting


Topic: Reaction Quotient for Electrochemical Cell  (Read 5661 times)

0 Members and 1 Guest are viewing this topic.

Offline mistertaylor92

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Reaction Quotient for Electrochemical Cell
« on: March 02, 2012, 06:52:19 PM »
For a lab we constructed an electrochemical cell that consists of two compartments; one consisting of a zinc electrode submersed in a 0.15 M zinc chloride solution (the anode) and another consisting of a platinum electrode submersed in a solution that is 0.5 M in both potassium ferrocyanide (K4Fe(CN)6•3H2O) and potassium ferricyanide (K3Fe(CN)6)  (the cathode). I need to calculate Q for the reaction.

From what I understand, the representation of the cell is as follows:

Zn(s) | Zn2+aq(0.15M) || Fe(CN)63-(aq)(0.10 M), Fe(CN)64-(aq)(0.10 M) | Pt

And the overall reaction is as follows:

Zn(s) + 2Fe(CN)64-(aq)  :rarrow:  Zn2+aq + 2Fe(CN)63-(aq)

Am I right to say that the reaction quotient for this reaction is equal to:

Q = [Zn2+]*[Fe(CN)63-]2/[Fe(CN)64-]2
   = (0.15 M)(0.10 M)2/(0.10 M)2
   = 0.15

The lab manual states that activities can be ignored. Thanks in advance.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27652
  • Mole Snacks: +1800/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Reaction Quotient for Electrochemical Cell
« Reply #1 on: March 03, 2012, 04:56:13 AM »
In general looks good, but why 0.1 if 0.5?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline mistertaylor92

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Reaction Quotient for Electrochemical Cell
« Reply #2 on: March 03, 2012, 06:02:05 AM »
My apologies that was a typo. The molarity of the iron compounds are both 0.1 M.

Sponsored Links