Question

Calculate the voltage of the cell.

Answer 1

Answer:

$E_C_e_l_l=-0.511$

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)

$Ag_(_s_)~=Ag^+_(_a_q_)~+~e^-$     Eo = -0.799 V

$Cd^2^+_(_a_q_)~+~2e^-~ =~ Cd_(_s_)$  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:

$2Ag_(_s_)~+Cd^2^+_(_a_q_)~=~2Ag^+~+~ Cd_(_s_)$     Eo = -1.201

Next, using the nerst equation we can calculate the potential of the whole cell:

$E_C_e_l_l=E^{\circ}-\frac{0.0592}{n}~Log~Q$

$E_C_e_l_l=E^{\circ}-\frac{0.0592}{n}Log\frac{[Ag^+]^2}{[Cd^2^+]}$

Now, the problem is Q. To calculate the concentrations for Q we have to use the Ksp of AgBr to calculate the free $Ag^+$ concentration, so:

     AgBr(s) <=> Ag+    + Br-

I          ----            0          1.0

C        ----             +x        +x

E        ----               x       1.0+ x

$Ksp=\frac{Products}{reactives}$

$5x10^-^1^3=[x][1]$

$x=5x10^-^1^3$

With the concentration of $Ag^+$ and $Cd^+^2$ given by the problem we can calculate Q:

$Q=\frac{[5x10^-^1^3]^2}{[0.05]}=5x10^-^2^4$

Finally, with the nerst equation we can calculate the total voltage:

$E_C_e_l_l=-1.201}-\frac{0.0592}{2}Log[5x10^-^2^4]$

$E_C_e_l_l=-0.511$