Question

If $\Delta G^\circ$ of acidic solution of a half cell $MnO_4^-/MnO_2$ is -xF. Calculate value of x.

$\left(E_{MnO_4^-/Mn^{2+}}^\circ=1.5V, E_{MnO_2/Mn^{2+}}^\circ=1.25V\right)$

Answer 1

5

Answer 2

To calculate the value of x, we need to determine the standard Gibbs free energy change ($\Delta G^\circ$) for the half-cell reaction $MnO_4^-/MnO_2$ in acidic solution. The reaction can be written as: $MnO_4^- \rightarrow MnO_2$

First, let's write the given half-reactions and their corresponding $\Delta G^\circ$ values. The relationship between $\Delta G^\circ$ and standard electrode potential ($E^\circ$) is given by: $\Delta G^\circ = -nFE^\circ$ where $n$ is the number of electrons transferred, and $F$ is Faraday's constant.

1. For $E_{MnO_4^-/Mn^{2+}}^\circ = 1.5V$: The half-reaction is $MnO_4^- \rightarrow Mn^{2+}$. The oxidation state of Mn in $MnO_4^-$ is +7. The oxidation state of Mn in $Mn^{2+}$ is +2. The change in oxidation state is $7 - 2 = 5$. So, 5 electrons are involved. The balanced half-reaction in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (Reaction 1) Here, $n_1 = 5$ and $E_1^\circ = 1.5V$. $\Delta G_1^\circ = -n_1 F E_1^\circ = -5 \times F \times 1.5 = -7.5F$

2. For $E_{MnO_2/Mn^{2+}}^\circ = 1.25V$: The half-reaction is $MnO_2 \rightarrow Mn^{2+}$. The oxidation state of Mn in $MnO_2$ is +4. The oxidation state of Mn in $Mn^{2+}$ is +2. The change in oxidation state is $4 - 2 = 2$. So, 2 electrons are involved. The balanced half-reaction in acidic medium is: $MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$ (Reaction 2) Here, $n_2 = 2$ and $E_2^\circ = 1.25V$. $\Delta G_2^\circ = -n_2 F E_2^\circ = -2 \times F \times 1.25 = -2.5F$

3. For the target half-cell $MnO_4^-/MnO_2$: The half-reaction is $MnO_4^- \rightarrow MnO_2$. The oxidation state of Mn in $MnO_4^-$ is +7. The oxidation state of Mn in $MnO_2$ is +4. The change in oxidation state is $7 - 4 = 3$. So, 3 electrons are involved. The balanced half-reaction in acidic medium is: $MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$ (Reaction 3) Let this be $\Delta G_3^\circ$.

We can obtain Reaction 3 by combining Reaction 1 and Reaction 2. We want $MnO_4^-$ on the reactant side and $MnO_2$ on the product side. Reaction 1: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ To get $MnO_2$ on the product side, we need to reverse Reaction 2: $Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-$ (Reversed Reaction 2) The $\Delta G^\circ$ for the reversed reaction is the negative of the original $\Delta G^\circ$: $\Delta G_{2,reversed}^\circ = -\Delta G_2^\circ = -(-2.5F) = 2.5F$

Now, add Reaction 1 and Reversed Reaction 2: $(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O)$ $+$ $(Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-)$

$MnO_4^- + (8-4)H^+ + (5-2)e^- + (2-4)H_2O \rightarrow MnO_2$ $MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$ This is exactly Reaction 3.

Since Gibbs free energy is an additive property, $\Delta G_3^\circ$ can be calculated as: $\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_{2,reversed}^\circ$ $\Delta G_3^\circ = -7.5F + 2.5F$ $\Delta G_3^\circ = -5.0F$

The problem states that $\Delta G^\circ$ for this half-cell is -xF. Comparing $-xF$ with $-5.0F$: $-xF = -5.0F$ $x = 5.0$

The final answer is $\boxed{5}$.

Explanation: The standard Gibbs free energy change ($\Delta G^\circ$) is an additive property. We are given standard electrode potentials ($E^\circ$) for two half-reactions: $MnO_4^-/Mn^{2+}$ and $MnO_2/Mn^{2+}$. We need to find $\Delta G^\circ$ for the $MnO_4^-/MnO_2$ half-reaction. We convert each $E^\circ$ into $\Delta G^\circ$ using $\Delta G^\circ = -nFE^\circ$. Then, we manipulate the given half-reactions to obtain the desired half-reaction and sum their corresponding $\Delta G^\circ$ values.

1. For $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$, $\Delta G_1^\circ = -5F(1.5V) = -7.5F$.
2. For $MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$, $\Delta G_2^\circ = -2F(1.25V) = -2.5F$. To get $MnO_4^- \rightarrow MnO_2$, we subtract the second reaction from the first (or add the reverse of the second reaction). The target reaction is $MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$. $\Delta G^\circ_{target} = \Delta G_1^\circ - \Delta G_2^\circ = -7.5F - (-2.5F) = -7.5F + 2.5F = -5.0F$. Given $\Delta G^\circ = -xF$, therefore $x = 5.0$.