Consider the following reaction
$$MnO_4^- + 8H^+ + 5e^- \to Mn^{+2} + 4H_2O$$, $$E^\circ = 1.51$$ V
The quantity of electricity required in Faraday to reduce five moles of $$MnO_4^-$$ is ______

The given half-reaction is: $$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O, \quad E^\circ = 1.51 \text{ V}$$

In this reduction half-reaction, manganese is reduced from the +7 oxidation state in $$MnO_4^-$$ to the +2 oxidation state in $$Mn^{2+}$$. The change in oxidation state is $$7 - 2 = 5$$, which corresponds to the gain of 5 electrons per $$MnO_4^-$$ ion.

Since 1 Faraday of electricity corresponds to 1 mole of electrons ($$1 \text{ F} = 96485 \text{ C}$$), the reduction of 1 mole of $$MnO_4^-$$ requires 5 Faradays of electricity.

For 5 moles of $$MnO_4^-$$, the total quantity of electricity required is:

$$\text{Total Faradays} = 5 \text{ mol of } MnO_4^- \times 5 \text{ F per mol} = 25 \text{ F}$$

Therefore, 25 Faradays of electricity are needed to reduce 5 moles of $$MnO_4^-$$ to $$Mn^{2+}$$.

The answer is $$\boxed{25}$$.