Respuesta :
Answer:
[tex]E_C_e_l_l=-0.511[/tex]
Explanation:
First we have to use the description of the cell and keep in mind that the left part is the oxidation and the right part is the reduction, so:
Ag(s)|AgBr(s), NaBr(aq, 1.0 M)||CdCl2(aq, 0.050 M)|Cd(s)
[tex]Ag_(_s_)~=Ag^+_(_a_q_)~+~e^-[/tex] Eo = -0.799 V
[tex]Cd^2^+_(_a_q_)~+~2e^-~ =~ Cd_(_s_)[/tex] Eo = -0.402 V
Then we have to multiply the first semireaction by "2" to obtain the same amount of electrons and add the 2 semireactions, so:
[tex]2Ag_(_s_)~+Cd^2^+_(_a_q_)~=~2Ag^+~+~ Cd_(_s_)[/tex] Eo = -1.201
Next, using the nerst equation we can calculate the potential of the whole cell:
[tex]E_C_e_l_l=E^{\circ}-\frac{0.0592}{n}~Log~Q[/tex]
[tex]E_C_e_l_l=E^{\circ}-\frac{0.0592}{n}Log\frac{[Ag^+]^2}{[Cd^2^+]}[/tex]
Now, the problem is Q. To calculate the concentrations for Q we have to use the Ksp of AgBr to calculate the free [tex]Ag^+[/tex] concentration, so:
AgBr(s) <=> Ag+ + Br-
I ---- 0 1.0
C ---- +x +x
E ---- x 1.0+ x
[tex]Ksp=\frac{Products}{reactives}[/tex]
[tex]5x10^-^1^3=[x][1][/tex]
[tex]x=5x10^-^1^3[/tex]
With the concentration of [tex]Ag^+[/tex] and [tex]Cd^+^2[/tex] given by the problem we can calculate Q:
[tex]Q=\frac{[5x10^-^1^3]^2}{[0.05]}=5x10^-^2^4[/tex]
Finally, with the nerst equation we can calculate the total voltage:
[tex]E_C_e_l_l=-1.201}-\frac{0.0592}{2}Log[5x10^-^2^4][/tex]
[tex]E_C_e_l_l=-0.511[/tex]
