At 298 K, a 1 litre solution containing 10 mmol of Cr$$_2$$O$$_7^{2-}$$ and 100 mmol of Cr$$^{3+}$$ shows a pH of 3.0. Given: Cr$$_2$$O$$_7^{2-} \to$$ Cr$$^{3+}$$; E$$^0 = 1.330$$ V and $$\frac{2.303RT}{F} = 0.059$$ V. The potential for the half cell reaction is $$x \times 10^{-3}$$ V. The value of $$x$$ is _____.

We need to find the electrode potential for the half-cell reaction $$Cr_2O_7^{2-} \rightarrow Cr^{3+}$$ under the given conditions.

We begin by writing the balanced half-cell reaction: $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$.

This is a reduction half-reaction involving the transfer of 6 electrons ($$n = 6$$). The standard electrode potential is $$E^0 = 1.330$$ V.

Next, we state the Nernst equation. The Nernst equation relates the electrode potential to the concentrations of species involved: $$E = E^0 - \frac{2.303RT}{nF}\log Q$$, where $$Q$$ is the reaction quotient. For this half-reaction: $$Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$.

We identify the given values. Volume = 1 L, so concentrations equal mmol values in mol/L:
$$[Cr_2O_7^{2-}] = 10 \text{ mmol/L} = 0.01 \text{ M} = 10^{-2}$$ M
$$[Cr^{3+}] = 100 \text{ mmol/L} = 0.1 \text{ M} = 10^{-1}$$ M
pH = 3.0, so $$[H^+] = 10^{-3}$$ M
$$\frac{2.303RT}{F} = 0.059$$ V, so $$\frac{2.303RT}{nF} = \frac{0.059}{6}$$ V

Calculating the logarithm of the reaction quotient gives:
$$\log Q = \log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$
$$= \log\frac{(10^{-1})^2}{(10^{-2})(10^{-3})^{14}}$$
$$= \log\frac{10^{-2}}{10^{-2} \times 10^{-42}}$$
$$= \log\frac{10^{-2}}{10^{-44}}$$
$$= \log(10^{42}) = 42$$

Substituting into the Nernst equation yields:
$$E = 1.330 - \frac{0.059}{6} \times 42$$
$$E = 1.330 - 0.059 \times 7$$
$$E = 1.330 - 0.413$$
$$E = 0.917 \text{ V}$$

Therefore, expressed in the required form, $$E = 0.917 \text{ V} = 917 \times 10^{-3}$$ V, and hence $$x = \mathbf{917}$$.