To calculate the quantity of electricity required to reduce 12.3 grams of nitrobenzene to aniline with 50 percent current efficiency, we can follow these steps:

1. Understand the Chemical Reaction
- The reduction of nitrobenzene (C6H5NO2) to aniline (C6H5NH2) involves the transfer of 6 electrons.
- The balanced half-reaction is:
\[
C_6H_5NO_2 + 6 e^- + 6 H^+ \rightarrow C_6H_5NH_2 + 2 H_2O
\]

2. Calculate the Molar Mass
- Molar mass of nitrobenzene (C6H5NO2) = 123 g/mol
- Molar mass of aniline (C6H5NH2) = 93 g/mol

3. Determine Moles of Nitrobenzene
- Moles of nitrobenzene in 12.3 grams:
\[
\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{12.3 \text{ g}}{123 \text{ g/mol}} \approx 0.1 \text{ mol}
\]

4. Calculate Required Electrons
- Each mole of nitrobenzene requires 6 moles of electrons:
\[
\text{Electrons required} = 0.1 \text{ mol} \times 6 = 0.6 \text{ mol e}^-
\]

5. Convert Moles of Electrons to Coulombs
- 1 mole of electrons = 96485 C:
\[
\text{Total charge} = 0.6 \text{ mol} \times 96485 \text{ C/mol} \approx 57891 \text{ C}
\]

6. Adjust for Current Efficiency
- With 50% efficiency, the effective charge required is:
\[
\text{Effective charge} = \frac{57891 \text{ C}}{0.5} = 115782 \text{ C}
\]

Conclusion
- The quantity of electricity required to reduce 12.3 grams of nitrobenzene to aniline, considering 50% current efficiency, is approximately **115782 Coulombs**.