Question #d891d

1 Answer
Jul 20, 2017

The cell diagram

color(white)(-----)color(red)["Zn"_((s)) | "Zn"^(2+) || "Ag"^(1+) | "Ag"_((s))

tells us that the anode is on the left side of ||, and the cathode is to the right of ||.

Knowing this, we can figure out that color(red)["Zn"_((s)) is going to be the one that is oxidized while color(red)["Ag"^(+) is going to be the one that is reduced. So the redox reaction would look something like this:

color(white)(-----)color(red)["2Ag"^(+) + Zn_(s)rarr"2Ag"_((s)) + "Zn"^(2+)

Seeing that we were given the standard reduction potentials, where the potentials are measured at "298 K, 1 atm, and solutions at 1 M", we can now go ahead and calculate the standard cell potential using the following equation:

color(white)(------)color(blue)[E_"cell"^@ = E_"red"^@ + E_"ox"^@

color(white)(-----)

ul"Standard Reduction Potentials"

"Zn"^(2+) + 2e^"-" rarr "Zn"_((s))color(white)(---)E_"red"^@ = -0.76 V

"Ag"^(+) + 1e^"-" rarr "Ag"_((s))color(white)(---)E_"red"^@ = +0.80 V

Note: Since we were given standard reduction potentials, in order to get the standard oxidation potential (E_"ox"^@) for zinc, we would just reverse the sign for the E_"red"^@. Consequently, we would show the reaction for zinc as being oxidized.

So,

color(white)(--)E_"red"^@ = -0.76 Vcolor(white)(--)becomescolor(white)(--)E_"ox"^@ = +0.76 V

Solve for standard cell potential

color(white)(--)color(blue)[E_"cell"^@ = E_"red"^@+E_"ox"^@]->(+0.80)+(+0.76)=barul (|+1.56 V |)

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Now, we use something called the "Nernst equation" to calculate the non-standard cell potential, which I believe is what your question is asking.

color(white)(-----)color(blue)[E_"cell" = E_"cell"^@-(0.059)/(n)logQ]

Where

"n = number of moles of electrons transferred (2)"
"Q = reaction quotient" [["Zn"^(2+)]]/[["Ag"^+]^2]

color(white)(--)

Now, we would plugin where appropriate and solve(ignored units)

color(blue)[E_"cell" = E_"cell"^@-(0.059)/(n)logQ]

E_"cell" =(+1.56)-((0.059))/((2))log([[0.001]]/[0.1]^2)

E_"cell" =(+1.56)-(-0.0295)

E_"cell" =color(magenta)[+1.58 V

"Answer": color(magenta)[+1.58 V