The magnitude of the change in oxidising power of the $$MnO_4^- / Mn^{2+}$$ couple is $$x \times 10^{-4}$$ V if the $$H^+$$ concentration is decreased from 1 M to $$10^{-4}$$ M at 25°C. (Assume concentration of $$MnO_4^-$$ and $$Mn^{2+}$$ to be same on change in $$H^+$$ concentration). The value of x is ______. (Rounded off to the nearest integer)
[Given: $$\frac{2.303RT}{F} = 0.059$$]

The half-reaction for the $$MnO_4^-/Mn^{2+}$$ couple is:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

The Nernst equation for this half-cell is:

$$E = E^{\circ} + \frac{0.059}{5} \log \frac{[MnO_4^-][H^+]^8}{[Mn^{2+}]}$$

Since the concentrations of $$MnO_4^-$$ and $$Mn^{2+}$$ remain the same, the change in potential depends only on the change in $$[H^+]$$. The change in $$E$$ when $$[H^+]$$ decreases from 1 M to $$10^{-4}$$ M is:

$$\Delta E = \frac{0.059}{5} \times 8 \times \log \frac{[H^+]_{new}}{[H^+]_{old}}$$

$$\Delta E = \frac{0.059 \times 8}{5} \times \log \frac{10^{-4}}{1}$$

$$\Delta E = \frac{0.472}{5} \times (-4) = 0.0944 \times (-4) = -0.3776 \text{ V}$$

The magnitude of the change in oxidising power is $$0.3776$$ V $$= 3776 \times 10^{-4}$$ V.

Therefore, the value of $$x$$ is $$\textbf{3776}$$.