Question

14.__/15 pts) In a concentration cell, a voltage can be created when one compartment has a concentrated solution while the other has a dilute solution. • Let's build one out of the Cu?"(aq) + 2e → Cu(s) half reaction, using 1.00 M Cu?"(aq) in one half cell and 1.58 x 10 M Cu?"(aq) in the other half cell. A) Label the diagram below; identify the copper electrodes and the [Cu?") in each half-cell along with the anode, cathode, and direction of electron flow. Salt B) Find the voltage for this cell at 298 K. C) What will the cell voltage be at equilibrium?

Answer 1

Cathode clinole How anode tu the Luth let dish of é flow anodel lose & e to Cathode (gain of e) anode Cathode comapartment (Right) Cat² conc IM compartiment heft conc. (1.58 x 10 +2 CM m) ( why I have taken cut? = 1.5x 104 at anode side. because during proces at anode oxidation will take place due to which cut? will form of already cut? cone is more then more cuta will form make it undisturbed to at anode vide taken les conce of cut2) now Cu (aq)+ it ze 2e ai Carst +2 neanston Hall cell rexns at Cathode +2. w(s) Reduction +2 at anode Coxidation) cu overall cath (IM) cu (1.58x104 Ecell tCathode - Eanode 0.34 0.34 =O Scanned with CamScan Now CS

Nernst Ecall equation Escell 0.0591, long [Cum) froduct Cutz Reactant n= no. of electrons exchange in overa n reactions = 0 - 010591 1.58X104 2 IM E L 01059) log M) @og (1:58) + log1104)] -0.0591 [ 0.198 - 4 log (10) [0.198-4] Toy (10) = 1 = -0.02955 [-3.802) 2 2 -0.029 55 [0.198-4] Ecell - oo|1234 Ooll v of equilibrium AG-nf Eall AGE 7 MF & cell at equilibrium AG=0 AG=0 Ecell =0 the cell will be dead CS Scanned with CamScanner