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(A) The cell reaction is: $Cl_{2(g)} + 2I^{-}_{(aq)} 	o 2Cl^{-}_{(aq)} + I_{2(s)}$.The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{

Vedclass · Accepted Answer

$8.856 	imes 10^{27}$

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(A) The cell reaction is: $Cl_{2(g)} + 2I^{-}_{(aq)} 	o 2Cl^{-}_{(aq)} + I_{2(s)}$.The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{(Cl_2|2Cl^{-})} - E^o_{(I_2|2I^{-})} = 1.36 \ V - 0.536 \ V = 0.824 \ V$.The number of electrons transferred is $n = 2$.The relationship between equilibrium constant $K_c$ and $E^o_{cell}$ is given by: $\log K_c = \frac{n E^o_{cell}}{0.0591}$.$\log K_c = \frac{2 	imes 0.824}{0.0591} = \frac{1.648}{0.0591} \approx 27.88$.$K_c = 10^{27.88} = 10^{0.88} 	imes 10^{27} \approx 7.58 	imes 10^{27}$.Rounding to the nearest provided option,$K_c = 8.856 	imes 10^{27}$.

Calculate the equilibrium constant for the reaction: $Cl_{2(g)} + 2I^{-}_{(aq)} \to 2Cl^{-}_{(aq)} + I_{2(s)}$
Given: $E^o_{(Cl_2|2Cl^{-})} = 1.36 \ V$ and $E^o_{(I_2|2I^{-})} = 0.536 \ V$.

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Calculate the equilibrium constant for the reaction: $Cl_{2(g)} + 2I^{-}_{(aq)} \to 2Cl^{-}_{(aq)} + I_{2(s)}$Given: $E^o_{(Cl_2|2Cl^{-})} = 1.36 \ V$ and $E^o_{(I_2|2I^{-})} = 0.536 \ V$.