Map Experimental information: A Cu/CuSO4 reference electrode and a silver wire i

Question

Map Experimental information: A Cu/CuSO4 reference electrode and a silver wire indicator electrode are used in a titration of a chloride solution with silver nitrate. The voltage is measured at 1 mL increments as the titration proceeds. The solubility product (Ksp) of silver chloride can be determined from the titration of a known cocnentration of chloride At the equivalence point the relevant shorthand notation for the electrochemical cell is: Cu 2+ (o.100 M)ll g Agol(s)llAg e balanced electrochemical reaction is: Cu (s) 2Ag (aq) 2Ag (s) 2+ What is the overall cell potential given: Ag Ag E (V) 0.800 (aq) e Cu Cu EU 0.338 2e Number

Explanation / Answer

Solution:- E0cell = E0cathode - E0anode

Anode is of copper since it's standard reduction potential is less than silver standard reduction potential.

E0cell = 0.800 V - 0.338 V = 0.462 V

second part:- For this we will use the the nernst equation..

Ecell =  E0cell - (0.0592/2) log ([Cu2+]/[Ag+]2)

Ecell is given as 190. mV that is 0.190 V and [Cu+2] is given as 0.100 M

value of n is 2 since two electrons are transferred in the balanced equation.

0.190 = 0.462 - (0.0592/2) log([0.100]/[Ag+])

0.190 - 0.462 = - (0.0296) log([0.100]/[Ag+]2)

-0.272 = -0.0296 log([0.100]/[Ag+]2)

we have negative sign on both sides so it is canceled out..

0.272/0.0296 = log([0.100]/[Ag+]2)

9.19 = log([0.100]/[Ag+]2)

taking antilog..

([0.100]/[Ag+]2) = 109.19 = 1.55 x 109

[Ag+]2 = 0.100/(1.55 x 109)

[Ag+]2 = 6.45 x 10-11

it's mentioned that. Ksp = [Ag+]2

so, Ksp = 6.45 x 10-11

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