The correct value of cell potential in volt for the reaction that occurs when the following two half cells…

The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is $\begin{aligned} & \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^0=+1.33 \mathrm{~V} \\ & \end{aligned}$
  1. $+0.01 \mathrm{~V}$
  2. $+0.89 \mathrm{~V}$
  3. $+1.77 \mathrm{~V}$
  4. $+2.65 \mathrm{~V}$

Solution

$\begin{aligned} \mathrm{E}_{\text {cell }}^0 & =\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ & =1.33-(-0.44) \\ & =1.33+0.44=+1.77 \mathrm{~V}\end{aligned}$

Asked in: NEET 2023 (Manipur)

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