Calculate the voltage of the cell.

ag(s)|agbr(s), nabr(aq, 1.0 m)||cdcl2(aq, 0.050 m)|cd(s)
agbr(s) + e- ⇌ ag(s) + br-
eo = 0.071 v
ag+ + e- ⇌ ag(s)
eo = 0.799 v
cd2+ + 2e- ⇌ cd(s)
eo = -0.402 v
ksp (agbr(s)) = 5.0 × 10-13

a. -0.678 v
b. -0.511 v
c. -0.423 v
d. 0.511 v

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)

    Eo = -0.799 V

 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:

    Eo = -1.201

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

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

     AgBr(s) <=> Ag+    + Br-

I          ----            0          1.0

C        ----             +x        +x

E        ----               x       1.0+ x

With the concentration of and given by the problem we can calculate Q:

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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2