Question

Find Cell Potential for: Zn(s) | [Zn+2] = 0.10M | NaOH =0.40M ; Kf = 1.0*10^6 || Cu(s)| [Cu+2] = 0.083 M I don't understand where the NaOH comes in and how to use the freezing point depression in this situation.

Answer 1

The reaction between Zn and Cu is

Cu²⁺ (Aq) + Zn (s) ----------- Cu(s) +Zn²⁺  (Aq)

Cu2+(aq)+2e-→Cu(s) E⁰red=+0.339 V (the values are Standard reduction values)

Zn(s)→Zn2+(aq)+2e^- E⁰ox=+0.762 V

E⁰ Cell =E_ox  + E_red

_{= 0.339+0.762 =1.101V}

E=E^o Cell + RT/nF * (log₁₀ a_ox /a_red ) R=Gas constant 8.314 J K^-1 mol^-1

  F=Faraday constant=96500 ,T=Temperature=298

E=E^o Cell + 0.0591/n * (log₁₀ a_ox /a_red ) n=number of moles=2

E =1.101+0.0591/2 * (log 0.10/0.083)

E= 1.101+ 0.02955 (0.093) =1.103

Just concentrate on the terms which are necessary to get the answer and determine the cell potential using Nernst equation