Question

The standard electrode potential for OCl-/Cl- and Cl-/1/2Cl2 respectively are 0.94V and -1.36V. Find E for OC-/1/2Cl2

Answer 1

Answer : The E for $OCl^-/\frac{1}{2}Cl_2$ is, -0.42 V

Explanation :

The half oxidation-reduction reactions are :

Reduction : $OCl^-+H_2O+2e^-\rightarrow Cl^-+2OH^-$   ......(1)

Oxidation : $Cl^-\rightarrow \frac{1}{2}Cl_2+e^-$      .......(2)

Now adding both equations, we get

$OCl^-+H_2O+e^-\rightarrow \frac{1}{2}Cl_2+2OH^-$

Now we have to calculate the standard cell potential for $OCl^-/\frac{1}{2}Cl_2$

$E^0_{[OCl^-/Cl^-]}=0.94V$

$E^0_{[Cl^-/\frac{1}{2}Cl_2]}=-1.36V$

Formula used :

$E^0=E^0_{reduction}+E^0_{oxidation}$

$E^0=E^0_{[OCl^-/Cl^-]}+E^0_{[Cl^-/\frac{1}{2}Cl_2]}$

$E^0=0.94V+(-1.36V)=-0.42V$

Therefore, the E for $OCl^-/\frac{1}{2}Cl_2$ is, -0.42 V

Answer 2

Given:

Standard electrode potential for OCl-/Cl-=0.94 V

Standard electrode potential for Cl-/1/2 Cl2=-1.36 V

To find:

$E^{0}$ for OCl-/1/2Cl2

Solution:

Reduction -half cell reaction:

$OCl^{-}+H_2O+2e-\rightarrow Cl^{-}+2OH^{-}$

Oxidation-half cell reaction:

$Cl^{-}\rightarrow 1/2Cl_2+e-$

Overall reaction:

$OCl^{-}+H_2O+e-\rightarrow1/2 Cl_2+2OH^{-}$

$E^{0}_{cathode}=0.94 V$

$E^{0}_{Anode}=-1.36V$

We know that

$E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{Anode}$

Using the formula

$E^{0}_{ OCl-/1/2Cl2}=0.94-(-1.36)=2.3 V$