# Question d891d

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 $\text{298 K, 1 atm, and solutions at 1 M} ,$ 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"^@

$\textcolor{w h i t e}{- - - - -}$

$\underline{\text{Standard Reduction Potentials}}$

${\text{Zn"^(2+) + 2e^"-" rarr "Zn"_((s))color(white)(---)E_"red}}^{\circ} = - 0.76 V$

${\text{Ag"^(+) + 1e^"-" rarr "Ag"_((s))color(white)(---)E_"red}}^{\circ} = + 0.80 V$

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 ${E}_{\text{red}}^{\circ}$. Consequently, we would show the reaction for zinc as being oxidized.

So,

$\textcolor{w h i t e}{- -} {E}_{\text{red}}^{\circ} = - 0.76 V \textcolor{w h i t e}{- -}$becomes$\textcolor{w h i t e}{- -} {E}_{\text{ox}}^{\circ} = + 0.76 V$

Solve for standard cell potential

$\textcolor{w h i t e}{- -} \textcolor{b l u e}{{E}_{\text{cell"^@ = E_"red"^@+E_"ox}}^{\circ}} \to \left(+ 0.80\right) + \left(+ 0.76\right) = \overline{\underline{| + 1.56 V |}}$

$- - - - - - - - - - - - - - - - - - - - -$

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

$\textcolor{w h i t e}{- - - - -} \textcolor{b l u e}{{E}_{\text{cell" = E_"cell}}^{\circ} - \frac{0.059}{n} \log Q}$

Where

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

$\textcolor{w h i t e}{- -}$

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

$\textcolor{b l u e}{{E}_{\text{cell" = E_"cell}}^{\circ} - \frac{0.059}{n} \log Q}$

${E}_{\text{cell}} = \left(+ 1.56\right) - \frac{\left(0.059\right)}{\left(2\right)} \log \left(\frac{\left[0.001\right]}{0.1} ^ 2\right)$

${E}_{\text{cell}} = \left(+ 1.56\right) - \left(- 0.0295\right)$

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

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