Question

A concentration cell is made up of the same materials in two halves of different concentrations. An example of a concentration cell is in the following drawing:

A. In the cathodic cell, nickel solution was added in concentration: [Ni+2] = 0.2M To measure the concentration of the solution in the anodic cell, measure the cell voltage and find it 0.009
volt.
A. Calculate the concentration of the unknown solution.
B. In another experiment, the half cell containing a nickel electrode, Ni immersed in nickel ion solution, Ni+2 in the unknown concentration was connected to a half cell containing platinum electrode, iron ion solution, Fe+2 at 0.3M concentration, and iron ion solution, Fe+3 concentration. 0.2M
Calculate the EMF at room temperature

Ni Salt bridge anode Ni cathode

Answer 1

The EMF of a cell is given by the expression, Ecathode - Eanode -(1) E = where , Ecathodes and e Ecathools and Eanod are the reduction potentials of Cathode and anode respectively, for a barticular reduction process, say A most ne - Ais : the a equation half - cell potential is as :- given by Nernst's En la = Egnt 1A - А"* / А ПЕ ГА**] for The half cell reaction (in terms of reduction) W; 2t/Ni cell is expressed on :- Ni tag) + de -Nico

- Griven E = 0.009 V [1:2] Cathocke = 0.2M let [N: 2+] anode = x M Then following equation <1) and (2) the EMF of the Concentration cell will be E = Ecathode - Eanoda ( ENIM RI s T och - ( Earn the on rearranging above equation becomes E = PT In (m. 21 ) cathoole [w: " Jonade QF

ting R = 8.314 Ik mol! F = 96500 C molt T=298k The above expression be E = 0.02955 log [N: 2+] Cathode to TN: 2+] anode. Putting the given date, we get 0.009 = 0.02955 lg. - log loe) = 0.3045 2 13. (a) : 1 1 3) - 341 . So, as at x = 0.1 This, the concentration of linknown solution [N; 27 Janode = 0.IM

in the two half ... The Cells half cell reactions can be written occurring as is Ni2+ + au Ni ; EN:atw. = -0.24v Feat té - Fu24 Een = 70.470 fou is the Enize Iwi = »:2-/N: - RT MF 2+ : -0.24 - F Note (w: 24 Tanode = Os M de calculated in (A) Janode 1 0.1 M in = is calculated Son Eui29 lei = -0.24 - 0.019 55 hores de 1 -0.26955 V

For Lily Efe37627 31/524 - R im Fe 31/727 – RT In [Fe2+] [Fe 3+ Putting values, we get FF 3+Fe2+ 0.77 - 0.0591 109 to 03 0.2 0.75 959 V This In E = Elegar / Excelsis - TO. 75959 – (-0.26955) Jv in 1.029 v = 1.03 v