Question

The electron equivalent per liter (\(e^{-}\)eq L\(^{-1}\)) of acetate ion solution of concentration 10 g L\(^{-1}\) is \(e^{-}\)eq L\(^{-1}\). (rounded off to one decimal place)

Hint:

Balance the oxidation of acetate to \(CO_2\) to find electrons released per mole, then multiply by the molar concentration (mass divided by molar mass 59 g/mol).

Answer 1

Correct Answer: 1.4

Step 1: Understanding the Question.
We need the electron equivalent (\(e^{-}\)eq) per litre of a 10 g L\(^{-1}\) solution of acetate ion (\(CH_3COO^{-}\)). An electron equivalent counts how many moles of electrons a substance can give up when it is fully oxidised. This idea is used to compare the "reducing power" of different organic substrates in a bioprocess or wastewater stream.

Step 2: Write the oxidation half reaction.
Acetate is oxidised all the way to carbon dioxide. Balancing carbon, hydrogen, oxygen and charge gives:
\[ CH_3COO^{-} + 2H_2O \rightarrow 2CO_2 + 7H^{+} + 8e^{-} \]
Check the balance: carbon 2 = 2, hydrogen \(3+4=7\) on the left equals \(7\) on the right, oxygen \(2+2=4\) equals \(2\times2=4\) on the right, and charge \(-1\) on the left equals \(7(+1)+8(-1)=-1\) on the right. So one mole of acetate ion releases exactly 8 moles of electrons on complete oxidation, so its electron equivalence is 8 eq per mole.

Step 3: Convert the given mass concentration to molar concentration.
The molar mass of the acetate ion \(CH_3COO^{-}\) (\(C_2H_3O_2^{-}\)) is
\[ M = 2(12) + 3(1) + 2(16) = 24 + 3 + 32 = 59 \ \text{g mol}^{-1} \]
So the molar concentration of the 10 g L\(^{-1}\) solution is
\[ C = \frac{10}{59} = 0.1695 \ \text{mol L}^{-1} \]

Step 4: Multiply by the electron equivalence per mole.
Since each mole gives 8 equivalents of electrons,
\[ \text{e}^{-}\text{eq L}^{-1} = 0.1695 \times 8 = 1.356 \ e^{-}\text{eq L}^{-1} \]

Final Answer:
Rounded off to one decimal place, the electron equivalent per litre of the acetate solution is
\[ \boxed{1.4 \ e^{-}\text{eq L}^{-1}} \]