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concentration cell :

e –. Ni bar B. Ni bar A. [Ni 2+ ] = 1.00 M. [Ni 2+ ] = 0.001 M. concentration cell :. different [ ]s of same species generating an emf. cathode. anode. salt bridge. (conc.). (dil.). -- E o for a [ ] cell =. 0.00 V. A. B. conc. dil. Bar A:. Bar B:. [X] dil. Add…. [X] conc.

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concentration cell :

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  1. e– Ni bar B Ni bar A [Ni2+] = 1.00 M [Ni2+] = 0.001 M concentration cell: different [ ]s of same species generating an emf cathode anode salt bridge (conc.) (dil.) -- Eo for a [ ] cell = 0.00 V

  2. A B conc. dil. Bar A: Bar B: [X] dil. Add… [X] conc. -- In the above example, cell will act to equalize [Ni2+]s, so… Ni(s)  Ni2+dil. + 2 e– Ni2+conc. + 2 e– Ni(s) Ni2+conc.  Ni2+dil. (for conc. cell only!) Thus, in general… Q = At equilibrium… [ ]s are = , so Q = 1 , and (by Nernst), E = 0 V.

  3. A [ ] cell has Cell A with [Cd2+] = 2.35 M and Cell B with [Cd2+] = 2.25 x 10–3 M. Identify anode and cathode, and calculate emf. Assume 298 K. [Cd2+] in B needs to increase, so Cd plate in B will be oxidized. Thus… B is anode and A is cathode. = 0.0894 V

  4. 0 K 0 0 K At equilibrium, DG = ___, E = ___ and Q = ___. The Nernst equation can be rearranged to give the relationship between K and Eo. – = log K = At 25oC (298 K): –DGo log K = 16.912 n Eo = 5706

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