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How many coulombs of electricity are req...

How many coulombs of electricity are required for the reduction of 1 mole of `MnO_(4) ^(-)` to `Mn^(2+)`?

A

96500 C

B

`9.65 xx l0^(6)` C

C

`4.83 xx10^(5)` C

D

`1.93 xx 10^(5)` C

Text Solution

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The correct Answer is:
To determine how many coulombs of electricity are required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \), we can follow these steps: ### Step 1: Determine the oxidation states First, we need to find the oxidation states of manganese in both \( \text{MnO}_4^{-} \) and \( \text{Mn}^{2+} \). - In \( \text{MnO}_4^{-} \): - Let the oxidation state of Mn be \( x \). - The oxidation state of oxygen is -2, and there are 4 oxygen atoms. - The overall charge of the ion is -1. Thus, the equation is: \[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \] - In \( \text{Mn}^{2+} \): - The oxidation state of Mn is +2. ### Step 2: Calculate the change in oxidation state Now, we can calculate the change in oxidation state: \[ \text{Change in oxidation state} = +7 - (+2) = +5 \] This indicates that manganese is reduced by gaining 5 electrons. ### Step 3: Relate moles of electrons to Faraday's constant Since 1 mole of \( \text{MnO}_4^{-} \) requires 5 moles of electrons for reduction, we can relate this to Faraday's constant. Faraday's constant (\( F \)) is approximately \( 96500 \, \text{C/mol} \). ### Step 4: Calculate total charge in coulombs To find the total charge (in coulombs) required for the reduction of 1 mole of \( \text{MnO}_4^{-} \): \[ \text{Total charge} = \text{moles of electrons} \times F = 5 \, \text{mol} \times 96500 \, \text{C/mol} \] Calculating this gives: \[ \text{Total charge} = 5 \times 96500 = 482500 \, \text{C} \] ### Step 5: Express the answer in scientific notation We can express \( 482500 \, \text{C} \) in scientific notation: \[ 482500 \, \text{C} = 4.825 \times 10^5 \, \text{C} \] ### Final Answer Thus, the total amount of electricity required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) is: \[ \boxed{4.83 \times 10^5 \, \text{C}} \]
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