# Given the half reactions Au+3 +3e- -> Au and Sn+4 +2e- -> Sn+2 in a galvanic cell, what is the voltage if [Au+3]=1.45M [Sn+4]=.50M and [Sn+2]=.87M?

Jan 12, 2017

$\textsf{{E}_{c e l l} = + 1.38 \textcolor{w h i t e}{x} V}$

#### Explanation:

The first thing to do is to calculate the emf of the cell under standard conditions.

We can do this using standard electrode potentials ($\textsf{{E}^{\circ}}$):

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

 " " "E"^@("V")

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$\textsf{S {n}^{4 +} \text{ "+" "2e" "rightleftharpoons" "Sn^(2+)" } + 0.15}$

$\textsf{A {u}^{3 +} \text{ "+" "3e" "rightleftharpoons" "Au" } + 1.52}$

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The more +ve half - cell will take in the electrons so the reactions in each half - cell will proceed in the direction shown by the arrows.

This gives the overall cell reaction:

$\textsf{2 A {u}^{3 +} + 3 S {n}^{2 +} \rightarrow 2 A u + 3 S {n}^{4 +}}$

To find $\textsf{{E}_{c e l l}^{\circ}}$ subtract the least +ve electrode potential from the most positive:

$\textsf{{E}_{c e l l}^{\circ} = + 1.52 - \left(+ 0.15\right) = 1.37 \textcolor{w h i t e}{x} V}$

Since we are not under standard conditions we now need to use The Nernst Equation.

A useful form of this at $\textsf{{25}^{\circ} C}$ is:

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

$\textsf{z}$ is the number of moles of electrons transferred which, in this case = 6.

$\textsf{Q}$ is the reaction quotient and is given by:

sf(Q=([Sn^(4+)]^(3))/([Au^(3+)]^(2)[Sn^(2+)]^(3))

$\therefore$$\textsf{Q = \frac{{\left(0.5\right)}^{2}}{{\left(1.45\right)}^{2} \times {\left(0.87\right)}^{3}} = 0.0902 \textcolor{w h i t e}{x} {\text{mol}}^{- 2} . {l}^{2}}$

$\therefore$$\textsf{{E}_{c e l l} = 1.37 - \frac{0.05916}{6} \log \left[0.0902\right]}$

$\textsf{{E}_{c e l l} = 1.37 + 0.103 = + 1.38 \textcolor{w h i t e}{x} V}$