1. **Problem Statement:** Calculate the cell potential for the reaction $$\mathrm{Cr_2O_7^{2-} + I^- \leftrightarrow Cr^{3+} + I_2}$$ under non-standard conditions with concentrations: $$[\mathrm{Cr^{3+}}] = 1.0 \times 10^{-5}$$, $$[\mathrm{Cr_2O_7^{2-}}] = 2.0$$, $$[\mathrm{H^+}] = 1.0$$, $$[\mathrm{I^-}] = 1.0$$. 2. **Balanced Redox Reaction:** $$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6I^- \leftrightarrow 2Cr^{3+} + 3I_2 + 7H_2O}$$ 3. **Reaction Quotient (Q):** $$Q = \frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}[\mathrm{I^-}]^6}$$ 4. **Substitute given concentrations:** $$Q = \frac{(1.0 \times 10^{-5})^2}{2.0 \times (1.0)^{14} \times (1.0)^6} = \frac{1.0 \times 10^{-10}}{2.0} = 5.0 \times 10^{-11}$$ 5. **Nernst Equation:** $$E = E^0 - \frac{0.0592}{n} \log Q$$ where $$E^0 = 0.796\,V$$ and $$n = 6$$ (number of electrons transferred). 6. **Calculate:** $$E = 0.796 - \frac{0.0592}{6} \log(5.0 \times 10^{-11})$$ $$= 0.796 - 0.009867 \log(5.0 \times 10^{-11})$$ 7. **Evaluate logarithm:** $$\log(5.0 \times 10^{-11}) = \log 5.0 + \log 10^{-11} = 0.6990 - 11 = -10.3010$$ 8. **Substitute back:** $$E = 0.796 - 0.009867 \times (-10.3010)$$ $$= 0.796 + 0.1016 = 0.8976\,V$$ 9. **Final answer:** $$\boxed{E_{cell} = 0.898\,V}$$