The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is

$$\begin{aligned} & \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \text { (aq) }+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}^{\circ}=+1.33 \mathrm{~V} \end{aligned}$$