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Step 1: Write the Half-Reaction and Determine the Number of Electrons
In acidic solution, permanganate is reduced to manganese(II) according to the balanced half-reaction:
MnO_4^- + 8H^+ + 5e^- \;\to\; Mn^{2+} + 4H_2O
From this, we see the number of electrons transferred, n , is 5.
Step 2: Write the Nernst Equation
The general form of the Nernst equation for this half-reaction at 25°C (where 2.303\,\frac{RT}{F} = 0.0591\text{ V} ) is:
E = E^\circ - \dfrac{0.0591}{n} \log\!\Biggl(\dfrac{[ \text{Products} ]}{[ \text{Reactants} ]}\Biggr)
For our purposes, we focus on the dependence on [H^+] while assuming [MnO_4^-] and [Mn^{2+}] remain constant and equal in the two scenarios.
Step 3: Express the Potential for [H+] = 1 M
When [H^+] = 1\,\text{M} , let this potential be E_1 :
E_1 = E^\circ \;-\; \dfrac{0.0591}{5} \log\!\Bigl(\dfrac{[Mn^{2+}]}{[MnO_4^-]}\bigl[1/(1)\bigr]^8\Bigr)
Since \bigl[1/(1)\bigr]^8 = 1 , the expression simplifies to:
E_1 = E^\circ \;-\; \dfrac{0.0591}{5} \log\!\Bigl(\dfrac{[Mn^{2+}]}{[MnO_4^-]}\Bigr)
Step 4: Express the Potential for [H+] = 10−4 M
When [H^+] = 10^{-4}\,\text{M} , let this potential be E_2 :
E_2 = E^\circ \;-\; \dfrac{0.0591}{5}\;\log\!\Bigl(\dfrac{[Mn^{2+}]}{[MnO_4^-]}\bigl[1/(10^{-4})\bigr]^8\Bigr)
Note that [1/(10^{-4})]^8 = (10^4)^8 = 10^{32} .
Step 5: Calculate the Magnitude of the Difference
We want the magnitude of E_1 - E_2 . Substituting the above expressions, the difference depends only on the change in [H^+] term:
\bigl|E_1 - E_2\bigr| \;=\; \dfrac{0.0591}{5} \;\log\!\bigl(10^{32}\bigr)
Since \log\!\bigl(10^{32}\bigr) = 32 , we get:
\bigl|E_1 - E_2\bigr| = \dfrac{0.0591}{5} \times 32
Calculate this:
\dfrac{0.0591 \times 32}{5} \;=\; 0.3776\;\text{V}
We can rewrite 0.3776\,\text{V} as 3776 \times 10^{-4}\,\text{V} .
Step 6: Identify the Value of x
The question defines the change in voltage as x \times 10^{-4}\,\text{V} . Since we have 0.3776\,\text{V} = 3776 \times 10^{-4}\,\text{V} , it follows that:
x = 3776
