Question

Consider the reaction corresponding to a voltaic cell and its standard cell potential. Zn(s) + Cu2+ (aq) Cu(s) + Zn2+ (aq) ЕО cell = 1.1032 V What is the cell potential for a cell with a 2.440 M solution of Zn2+ (aq) and 0.1636 M solution of Cu²+ (aq) at 439.5 K?

Answer 1

F is faraday number

* Keq is equilibrium constant which is written as the ration pr product of concetration of product to the product of concetration of Reactant .

This concentration is written 1 for pure solid.

* n is number of Electron transferred in given Reaction.

2+ Zn (S) 2+ + Cu (0) a Curs) & Zn coa) E cell =1:1032V [Zn2+ 2.44 M 2 3 4 5 2+ cau 427) - 0.1636M T = 4 39.5K Zn cs) 2t Zn 72e Number of electrons transferred = 2=n To findout cell betential [Ewell), herst equation is used – Nernst equateen E - Eiell P RT Keg NE R = 0.314 Ilma-k n=2 Coulomb T = 439,5K fa 9 6500 mole Putting values 8.314 X 439.5 [Zn2+) Eceli 1.1032 - en 2 x 96500 [42+] - 1.1032 - 7.44 Oro 189 en 420) 0.1636 -110 32-0.05116 E cell = l.os2 velt Answer CS Scanned with CamScanner