The magnitude of the change in oxidising power of the MnO4- / Mn2+ couple is x * 10^-4 V, if the H+

Q: Why is the coefficient of $\text{H}^+$ exactly 8?

To balance the 4 oxygen atoms in $\text{MnO}_4^-$, we need 4 $\text{H}_2\text{O}$ molecules on the product side. This introduces 8 hydrogen atoms, which must be balanced by 8 $\text{H}^+$ ions on the reactant side.

Q: Why do we assume the concentrations of $\text{MnO}_4^-$ and $\text{Mn}^{2+}$ remain the same?

The problem explicitly states this assumption to isolate the effect of pH on the electrode potential. In a real scenario, changing pH might shift the equilibrium, but here we are calculating the instantaneous potential change.

Q: Why does a decrease in $\text{H}^+$ concentration lower the oxidising power?

According to Le Chatelier's principle, decreasing the concentration of a reactant ($\text{H}^+$) shifts the equilibrium to the left, making the reduction less favorable. This is mathematically reflected in a lower (less positive) reduction potential.

Q: What happens to the standard potential $E^\circ$ during this change?

The standard potential $E^\circ$ is a constant value defined at standard conditions ($1\text{ M}$ concentration for all aqueous species, $298\text{ K}$). It does not change; only the actual cell potential $E$ changes with concentration.

Q: Why do we use 0.059 instead of 0.0591?

The problem explicitly provides the value $\frac{2.303 RT}{F} = 0.059$. Always use the constants provided in the question to avoid rounding errors in integer-type questions.

Question

The magnitude of the change in oxidising power of the $	ext{MnO}_4^- / 	ext{Mn}^{2+}$ couple is $x 	imes 10^{-4}	ext{ V}$, if the $	ext{H}^+$ concentration is decreased from $1	ext{ M}$ to $10^{-4}	ext{ M}$ at $25^\circ	ext{C}$. (Assume concentration of $	ext{MnO}_4^-$ and $	ext{Mn}^{2+}$ to be same on change in $	ext{H}^+$ concentration). The value of $x$ is ......... (Rounded off to the nearest integer). $\left[	ext{Given, } \frac{2.303 RT}{F} = 0.059ight]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a beaker containing a vibrant purple solution of permanganate ions ($	ext{MnO}_4^-$) ready to snatch electrons and become the pale pink manganese ion ($	ext{Mn}^{2+}$). This is a classic redox couple, and its ability to pull electrons—its **oxidising power**—is heavily dependent on the environment.

Specifically, this reaction doesn't just need electrons; it is incredibly thirsty for protons ($	ext{H}^+$).

The balanced half-cell reaction reveals this dependency beautifully:

$	ext{MnO}_4^- + 8	ext{H}^+ + 5	ext{e}^- \longrightarrow 	ext{Mn}^{2+} + 4	ext{H}_2	ext{O}$

Notice that massive stoichiometric coefficient of $8$ in front of the hydrogen ions. This tells us that the pH of the solution is going to play a dominant role in determining the electrode potential.

### The Master Equation

To quantify this oxidising power, we bring in our ultimate tool: the **Nernst Equation**.

It connects the standard potential to the actual concentrations in our beaker. For this half-cell, the equation takes the form:

$E = E^\circ - \frac{0.059}{n} \log \frac{[	ext{Mn}^{2+}]}{[	ext{MnO}_4^-][	ext{H}^+]^8}$

Here, the number of electrons transferred, $n$, is $5$. The concentration of $	ext{H}^+$ is raised to the power of $8$, which means even a tiny change in pH will be amplified exponentially in the reaction quotient.

### The Initial and Final States

Let's look at our starting point. Initially, the solution is highly acidic with $[	ext{H}^+] = 1	ext{ M}$.

Plugging this into our equation, the $	ext{H}^+$ term in the denominator simply becomes $1^8$, which is just $1$.

$E_1 = E^\circ - \frac{0.059}{5} \log \frac{[	ext{Mn}^{2+}]}{[	ext{MnO}_4^-]}$

Now, we drastically reduce the acidity, dropping the $	ext{H}^+$ concentration to $10^{-4}	ext{ M}$.

Substituting this new value, the denominator now contains $(10^{-4})^8$, which evaluates to $10^{-32}$.

$E_2 = E^\circ - \frac{0.059}{5} \log \frac{[	ext{Mn}^{2+}]}{[	ext{MnO}_4^-](10^{-32})}$

### Final Calculation

We can elegantly rewrite $E_2$ by pulling that $10^{-32}$ out of the denominator:

$E_2 = E_1 - \frac{0.059}{5} \log(10^{32})$

The problem asks for the magnitude of the change in oxidising power, which is simply the absolute difference between $E_1$ and $E_2$.

$|E_1 - E_2| = \frac{0.059}{5} 	imes 32$

Calculating this gives us:

$\Delta E = 0.3776	ext{ V}$

To match the requested format of $x 	imes 10^{-4}	ext{ V}$, we shift the decimal point four places to the right:

$\Delta E = 3776 	imes 10^{-4}	ext{ V}$

Thus, our final integer value for $x$ is **$3776$**. The dramatic drop in oxidising power perfectly illustrates Le Chatelier's principle: starving the reaction of protons makes it much harder for permanganate to act as an oxidising agent.

Enter Numerical Value:

Visualized Solution

\text{Half-Cell Reaction}

\text{Nernst Equation}

\text{Initial State } (E_1)

\text{Final State } (E_2)

\text{Simplifying } E_2

\text{Change in Potential } (\Delta E)

\text{Final Answer}

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

The Initial and Final States

Final Calculation

Similar Questions

The reduction potential ( $E^{0}$ , in V) of $MnO_{4}^{-} (aq) / Mn (s)$ is _______. [Given : $E_{(MnO_{4}^{-} (aq) / MnO_{2} (s))}^{0} = 1.68 V$ ; $E_{(MnO_{2} (s) / Mn^{2 +} (aq))}^{0} = 1.21 V$ ; $E_{(Mn^{2 +} (aq) / Mn (s))}^{0} = - 1.03 V$ ]

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

What would be the electrode potential for the given half-cell reaction at $pH = 5$ ? …… . $2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}$ ; $E_{red}^{\circ} = 1.23 V$ ( $R = 8.314 J mol^{- 1} K^{- 1}$ ; Temp $= 298 K$ ; oxygen under std. atm. pressure of $1 bar$ )

The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction is ( $F = 96500 C mol^{- 1}$ )