# Consider the reaction 2AgCl+Cu^(2+)rarr2AgCl+Cu. Given that DeltaG_("rxn")=-22.kJ, How do you find the value for E_(rxn)^@ ?

Aug 3, 2017

You can do it like this:

#### Explanation:

The expression you need is :

$\textsf{\Delta {G}^{\circ} = - n F {E}_{r x n}^{\circ}}$

Where n is the no. of moles of electrons transferred which, in this case = 2.

$\textsf{F}$ is the charge on 1 mole of electrons and is the Faraday Constant which = $\textsf{9.65 \times {10}^{4} \textcolor{w h i t e}{x} \text{C/mol}}$

$\textsf{\Delta {G}^{\circ} = - n F {E}_{r x n}^{\circ}}$

$\therefore$$\textsf{- 22.2 \times {10}^{3} = - 2 \times 9.65 \times {10}^{4} \times {E}_{r x n}^{\circ}}$

$\textsf{{E}_{r x n}^{\circ} = \frac{- 22.2 \times {10}^{3}}{- 2 \times 9.65 \times {10}^{4}} = + 0.115 \textcolor{w h i t e}{x} V}$

We can check this by comparing standard electrode potentials which can be looked up:

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)

$\textsf{A g C {l}_{\left(s\right)} + e r i g h t \le f t h a r p \infty n s A {g}_{\left(s\right)} + C {l}_{\left(a q\right)}^{-} \text{ } {E}^{\circ} = + 0.2223 \textcolor{w h i t e}{x} V}$

$\textsf{C {u}_{\left(a q\right)}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C {u}_{\left(s\right)} \text{ } {E}^{\circ} = + 0.34 \textcolor{w h i t e}{x} V}$

stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)

The most +ve half cell will take in the electrons so the half - reactions will proceed in the directions shown by the arrows.

$\textsf{{E}_{r x n}^{\circ}}$ is the arithmetic difference between the two 1/2 cells which you get by subtracting the least +ve $\textsf{{E}^{\circ}}$ from the most +ve $\Rightarrow$

$\textsf{{E}_{r x n}^{\circ} = + 0.34 - \left(+ 0.2223\right) = + 0.1177 \textcolor{w h i t e}{x} V}$

This agrees ok with the calculated value.