Question

A chemist designs a galvanic cell that uses these two half-reactions:

Half-reaction Standard reduction potential
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) E⁰ red = +1.23 V
Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E⁰ red = +0.771 V

(a) Write a balanced equation for the half-reaction that happens at the cathode.
(b) Write a balanced equation for the half-reaction that happens at the anode.
(c) Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written.
(d) Do you have enough information to calculate the cell voltage under standard conditions?

Answer 1

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$  

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$  

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$     $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$     $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$     $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$     $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.