Question

Given a half-reaction Cu2+(aq) + 2e- --> Cu(s) with an electrode potential of +0.34V and a half-reaction Fe3+(aq) + e- --> Fe2+(aq) with an electrode potential of +0.77V, what is the cell potential for the electrochemical reaction when Cu2+(aq) and Fe3+(aq) ions react to form Cu(s) and Fe2+(aq) ions?

Answer 1

To determine the cell potential for the electrochemical reaction, we first need to balance the half-reactions and then combine them to form the overall redox reaction.The given half-reactions are:1. Cu2+ aq  + 2e- --> Cu s  with an electrode potential of +0.34V2. Fe3+ aq  + e- --> Fe2+ aq  with an electrode potential of +0.77VTo balance the number of electrons in both half-reactions, we need to multiply the second half-reaction by 2:2 Fe3+ aq  + e- --> Fe2+ aq   becomes 2Fe3+ aq  + 2e- --> 2Fe2+ aq  with an electrode potential of +0.77VNow, we can combine the two half-reactions:Cu2+ aq  + 2e- + 2Fe3+ aq  + 2e- --> Cu s  + 2Fe2+ aq The overall redox reaction is:Cu2+ aq  + 2Fe3+ aq  --> Cu s  + 2Fe2+ aq To find the cell potential, we subtract the electrode potential of the half-reaction with the lower potential from the one with the higher potential:E_cell = E_higher - E_lowerE_cell = 0.77V - 0.34VE_cell = 0.43VSo, the cell potential for the electrochemical reaction when Cu2+ aq  and Fe3+ aq  ions react to form Cu s  and Fe2+ aq  ions is 0.43V.