E° of Fe3 /Fe2 system is 0-77 v. In the presence of phosphoric acid,Fe3 is removed as phosphate complex as shown in below : Fe3 HgPO4= Fe(H2PO4)2 What will be the change on the potential of the system? If the value of equilibrium constant of the above equilibrium is 1x103, calculate the approximate value of the potential of the Fe3 /Fe2 s



Answer :

[tex]\[ E = E^\circ - \frac{RT}{nF} \ln \left( \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \right) \]\\2. Given standard electrode potential: \[ E^\circ = 0.77 \, \text{V} \]3. Equilibrium constant for the complexation reaction: \[ K = \frac{[\text{Fe}(\text{H}_2\text{PO}_4)_2]}{[\text{Fe}^{3+}][\text{HgPO}_4]} \]4. New concentration of Fe3+ after complex formation: \[ [\text{Fe}^{3+}]_{\text{new}}[/tex][tex]\[ E = 0.77 \, \text{V} - \frac{RT}{F} \ln \left( \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]_{\text{new}}} \right) \]6. Final potential: \[ E \approx 0.5925 \, \text{V} \][/tex]