Question

State Faraday’s first law of electrolysis. How much electricity, in terms of Faraday, is required to reduce one mole of MnO4− to Mn2+ ion?

Answer 1

Faraday’s First Law of Electrolysis

Faraday’s first law of electrolysis states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

Mathematical expression:

\[ m = \frac{M \cdot I \cdot t}{F} \]

Where:

• \(m\) = mass of the substance deposited (in grams)
• \(M\) = molar mass of the substance (g/mol)
• \(I\) = current (in amperes)
• \(t\) = time (in seconds)
• \(F\) = Faraday’s constant = 96,485 C/mol

Application: Reduction of \({MnO_4^-}\) to \({Mn^{2+}}\)

In this reaction, manganese's oxidation state changes from +7 to +2. This requires the following number of electrons for reduction:

\[ {Mn^{7+} -> Mn^{2+}} \quad \Rightarrow \quad 5 \text{ electrons} \]

Consequently, the charge \(Q\) required to reduce 1 mole of \({MnO_4^-}\) to \({Mn^{2+}}\) is calculated as:

\[ Q = 5 \times F = 5 \times 96,\!485 = 482,\!425 \text{ C} \]

Therefore, 482,425 C of electricity is necessary to reduce 1 mole of \({MnO_4^-}\) to \({Mn^{2+}}\).