# Question f7401

May 22, 2016

$\text{K} \cong {10}^{62}$ which can be regarded as infinitely large.

#### Explanation:

Arrange the ${\text{E}}^{\circ}$ values in increasing order:

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

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$F {e}_{\left(a q\right)}^{3 +} + e r i g h t \le f t h a r p \infty n s F {e}_{\left(a q\right)}^{2 +} \text{ } + 0.77$

$M n {O}_{4 \left(a q\right)}^{-} + 8 {H}_{\left(a q\right)}^{+} + 5 e r i g h t \le f t h a r p \infty n s M {n}_{\left(a q\right)}^{2 +} + 4 {H}_{2} {O}_{\left(l\right)} \text{ } + 1.51$

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The Mn(VII) 1/2 cell has the more +ve value so the reaction will be driven in the direction indicated by the arrows

To find ${\text{E}}_{c e l l}^{\circ}$ you subtract the least positive value from the most +ve:

$\text{E"_(cell)^(@)=1.51-0.77=+0.74"V}$.

The Nernst Equation at ${25}^{\circ} \text{C}$ can be written:

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

$\text{Q}$ is the reaction quotient

$n$ is the number of moles of electrons transferred, which in this case $= 5$

As the reaction proceeds, the potential difference between the two 1/2 cells falls.

When it reaches zero the reaction has reached equilibrium so now we can write:

$0 = {E}_{c e l l}^{\circ} - \frac{0.05916}{5} \log \text{K}$

Note that $\text{Q}$ has been replaced by $\text{K}$, the equilibrium constant.

Putting in the value for ${E}_{c e l l}^{\circ}$ and rearranging $\Rightarrow$

$\log \text{K} = \frac{0.74 \times 5}{0.05916} = 62.5$

$\therefore \text{K} = 3.48 \times {10}^{62}$

This is a number so large that we can say that the reaction has gone to completion.