Chapter 20 - Electrochemistry - Example 20-3 - Combining E Values into E(cell) for a Reaction - Page 873: Practice Example A

0.587 V

Oxidation: $2Fe^{2+}\rightarrow 2Fe^{3+}+2e^{-}$ Reduction: $Cl_{2}+2e^{-}\rightarrow 2Cl^{-}$ $E^{\circ}_{cell}=E^{\circ}(\text{reduction half-cell})-E^{\circ}(\text{oxidation half-cell})$ $=1.358\,V-0.771\,V=0.587\,V$