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Standard reduction potential of MnO4-/MnO2 couple
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Lostinspace1205
Standard reduction potential of MnO4-/MnO2 couple
The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.
Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.
The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.
Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.
Well I still have a question - The balanced cell reaction would $2$ times $(1) - $ $5$ times of $(2)$ and thus in total 10 $e^{-1}$ are used, So shouldnt we multiply the gibbs energy by them and after subtracting divide by 10? Why would that be wrong?More
Can you calculate what the resultant chemical equation is if you take $2\times (1) - 5 \times (2)$? Is that what you are trying to find $\Delta G^\circ$ for?More
The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.
Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
$$\begin{align} \ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\ \ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2} \end{align}$$
The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:
$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
$$\begin{align} \Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\ &= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\ &= 491.11~\mathrm{kJ~mol^{-1}} \\ E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\ &= +1.697~\mathrm{V} \end{align}$$
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.
The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.
Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
$$\begin{align}\ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\\ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}\end{align}$$
The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:
$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
$$\begin{align}\Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\&= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\&= 491.11~\mathrm{kJ~mol^{-1}} \\E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\&= +1.697~\mathrm{V}\end{align}$$
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.
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