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I am referring to the following three reactions. Two of these reactions are imaginary (M$\ce{M}$ and X$\ce{X}$ are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$$$\begin{alignat}{3} \ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\ \ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\ \ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)} \end{alignat}$$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a)$\text{(a)}$ and (c)$\text{(c)}$;

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$$$ \begin{align} &{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\ &E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\ &\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b)$\text{(b)}$ and (c)$\text{(c)}$;

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$$$ \begin{align} &{-}\text{(b)} \times 5 + \text{(c)}: \\ \ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ &E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\ &\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F \end{align}$$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$$E^\circ$ values, reaction (2)$(2)$ is more feasible than reaction (1)$(1)$. Am I correct  ?
  3. According to $\Delta G^{\circ}$$\Delta G^\circ$ values, reaction (1)$(1)$ is more feasible than reaction (2)$(2)$. Am I correct  ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions  ? Should we always resort to $\Delta G$  ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict  ?
  6. If not how can these results be explained  ?

I am referring to the following three reactions. Two of these reactions are imaginary (M and X are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a) and (c);

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b) and (c);

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$ values, reaction (2) is more feasible than reaction (1). Am I correct  ?
  3. According to $\Delta G^{\circ}$ values, reaction (1) is more feasible than reaction (2). Am I correct  ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions  ? Should we always resort to $\Delta G$  ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict  ?
  6. If not how can these results be explained  ?

I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).

$$\begin{alignat}{3} \ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\ \ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\ \ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)} \end{alignat}$$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations $\text{(a)}$ and $\text{(c)}$;

$$ \begin{align} &{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\ &E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\ &\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations $\text{(b)}$ and $\text{(c)}$;

$$ \begin{align} &{-}\text{(b)} \times 5 + \text{(c)}: \\ \ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ &E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\ &\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F \end{align}$$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
  3. According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
  6. If not how can these results be explained?
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I am referring to the following three reactions. Two of these reactions are imaginary (M and X are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a) and (c);

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b) and (c);

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  3. According to $\Delta G^{\circ}$ values, reaction (1) is more feasible than reaction (2). Am I correct ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions ? Should we always resort to $\Delta G$ ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict ?
  6. If not how can these results be explained ?

Thanks a lot.

I am referring to the following three reactions. Two of these reactions are imaginary (M and X are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a) and (c);

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b) and (c);

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  3. According to $\Delta G^{\circ}$ values, reaction (1) is more feasible than reaction (2). Am I correct ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions ? Should we always resort to $\Delta G$ ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict ?
  6. If not how can these results be explained ?

Thanks a lot.

I am referring to the following three reactions. Two of these reactions are imaginary (M and X are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a) and (c);

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b) and (c);

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  3. According to $\Delta G^{\circ}$ values, reaction (1) is more feasible than reaction (2). Am I correct ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions ? Should we always resort to $\Delta G$ ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict ?
  6. If not how can these results be explained ?
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Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?

I am referring to the following three reactions. Two of these reactions are imaginary (M and X are imaginary).

$$ \begin{align} \ce{X- + 2e -> X^{3-} \qquad &E^{\circ}_{X^{3-}/X^{-}}=\ 0.9V \ &---(a) }\\ \ce{2H+ + MO^{2+} + e -> M^{3+} + H2O \qquad &E^{\circ}_{{MO^{2+}}/{M^{3+}}}=\ 0.5V \ &---(b)\\ } \ce{8H+ + MnO^{-}_4 + 5e -> Mn^2+ + 4H2O \qquad &E^{\circ}_{{MnO^{-}_4}/{Mn^2+}}=\ 1.5V \ &---(c)} \end{align} $$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations (a) and (c);

$$ \begin{align} &(-a) \times 5 + (c) \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO^{-}_4 -> 2Mn^2+ + 5X^- + 8H2O \qquad &E^{\circ}_1 &\ ---(1)} \\ \\ &E^{\circ}_1 = 1.5 + (-0.9) = 0.6 V \\ &\Delta G^{\circ}_1 = -nE^{\circ}_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations (b) and (c);

$$ \begin{align} &(-b) \times 5 + (c): \\ \ce{&5M^3+ + H2O + MnO^{-}_4 -> Mn^{2+} + 5MO^2+ + 2H+ \qquad &E^{\circ}_2 &\ ---(2)}\\ \\ &E^{\circ}_2 = 1.5 + (-0.5) = 1.0 V \\ &\Delta G^{\circ}_2 = -nE^{\circ}_2F = -5 \times 1.0 \times F = -5F \end{align} $$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^{\circ}$ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  3. According to $\Delta G^{\circ}$ values, reaction (1) is more feasible than reaction (2). Am I correct ?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions ? Should we always resort to $\Delta G$ ?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict ?
  6. If not how can these results be explained ?

Thanks a lot.