Cr_2O_7^{2-} + I^{-} to I_2 + Cr^{3+}E^0_{cel | vedclass

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(D) The given reaction is: $Cr_2O_7^{2-} + I^{-} 	o I_2 + Cr^{3+}$In this reaction,$I^{-}$ is oxidized to $I_2$ (anode) and $Cr_2O_7^{2-}$ is reduced

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$0.54$

Vedclass · Answer

(D) The given reaction is: $Cr_2O_7^{2-} + I^{-} 	o I_2 + Cr^{3+}$In this reaction,$I^{-}$ is oxidized to $I_2$ (anode) and $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$ (cathode).The standard cell potential is given by: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$Substituting the values: $0.79 \ V = 1.33 \ V - E^0_{I_2}$Rearranging for $E^0_{I_2}$: $E^0_{I_2} = 1.33 \ V - 0.79 \ V = 0.54 \ V$.Therefore,the correct option is $(D)$.

$Cr_2O_7^{2-} + I^{-} \to I_2 + Cr^{3+}$
$E^0_{cell} = 0.79 \ V$
$E^0_{Cr_2O_7^{2-}} = 1.33 \ V$,$E^0_{I_2}$ is ........... $V$

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$Cr_2O_7^{2-} + I^{-} \to I_2 + Cr^{3+}$$E^0_{cell} = 0.79 \ V$$E^0_{Cr_2O_7^{2-}} = 1.33 \ V$,$E^0_{I_2}$ is ........... $V$