# Consider the 2 half - cells : sf(Cu^(2+)+2erightleftharpoonsCu) sf(E^(@)=+0.34"V") sf(Ag^(+)+erightleftharpoonsAg) sf(E^(@)=+0.80"V") What is the solution to the points below?

## What is the standard emf of the cell formed from the two 1/2 cells ? What is the emf of the cell formed from the two half - cells if $\left[C {u}^{2 +}\right] = 0.130 M$ and $\left[A {g}^{+}\right] = 0.0001 M$ ?

Nov 25, 2016

$\textsf{{E}_{c e l l}^{\circ} = + 0.46 \textcolor{w h i t e}{x} V}$

$\textsf{{E}_{c e l l} = + 0.25 \textcolor{w h i t e}{x} V}$

#### Explanation:

List the 1/2 equations in order least positive to most positive:

$\textsf{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times x} {E}^{\circ} \textcolor{w h i t e}{\times} \left(V\right)}$

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)

$\textsf{C {u}^{2 +} \text{ "+" "2e" "rightleftharpoons" } C u \textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} + 0.34}$

$\textsf{A {g}^{+} \text{ "+" "e" "rightleftharpoons" } A g \textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times} + 0.80}$

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)

Note we use the $r i g h t \le f t h a r p \infty n s$ symbol to show that the 1/2 cells can go in either direction depending on what they are coupled with.

The 1/2 cell with the most +ve ${\text{E}}^{\circ}$ is the one which will take in the electrons.

From this we can see that the 2nd 1/2 cell will be driven left to right and the 1st 1/2 cell right to left in accordance with the arrows.

So the cell reaction when it is working is:

$\textsf{C u + 2 A {g}^{+} \rightarrow C {u}^{2 +} + 2 A g}$

Because $\textsf{{E}_{c e l l}^{\circ}}$ is an empirically measured quantity it always has a +ve value. So to calculate it just subtract the least +ve value from the most +ve:

$\textsf{{E}_{c e l l}^{\circ} = + 0.80 - \left(+ 0.34\right) = + 0.46 \textcolor{w h i t e}{x} V}$

To find the emf of the cell under non - standard condition we need to use the Nernst Equation. At $\textsf{{25}^{\circ} C}$ this simplifies down to:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{z} \log Q}$

$\textsf{Q}$ is the reaction quotient which, in this case is given by:

$\textsf{Q = \frac{\left[C {u}^{2 +}\right]}{{\left[A {g}^{+}\right]}^{2}}}$

$\textsf{z}$ is the no. moles of electrons transferred which, in this case = 2.

Putting in the numbers:

$\textsf{{E}_{c e l l} = + 0.46 - \frac{0.05916}{2} \log \left[\frac{0.130}{{\left({10}^{- 4}\right)}^{2}}\right] \text{ } V}$

$\textsf{{E}_{c e l l} = + 0.46 - \left(0.02958 \times 7.1139\right) \text{ } V}$

$\textsf{{E}_{c e l l} = + 0.46 - 0.2104 = 0.25 \textcolor{w h i t e}{x} V}$