Question

The redox reaction of peroxydisulfate with iodide has been used for many years as part of the iodine clock reaction. If E°cell= 1.587 V and E° of the cathode half-cell is 0.536 V, what is E° of the anode half-cell? S2O82–(aq) + 2H+ + 2I –(aq) → 2HSO4–(aq) + I2(aq)

Answer 1

Answer : The value of $E^0$  of anode half-cell is, -1.051 V

Explanation :

The given redox reaction is,

$S_2O_8^{2-}(aq)+2H^+(aq)+2I^-(aq)\rightarrow 2HSO_4^-(aq)+I_2(aq)$

From this we conclude that, the iodine (I) undergoes oxidation by loss of electrons and thus act as anode. Thiosulfate $(S_2O_8^{2-})$ undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $2I^-(aq)\rightarrow I_2(aq)+2e^-$     $E^0_{anode}=?$

Reaction at cathode (reduction) : $S_2O_8^{2-}(aq)+2H^++2e^-\rightarrow 2HSO_4^-(aq)$     $E^0_{cathode}=0.536V$

Now we have to calculate the $E^0_{anode}$  half-cell.

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

As we are given,

$E^0$  = 1.587 V

Now put all the given values in this relation, we get:

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

$1.587V=0.536V-E^o_{anode}$

$E^o_{anode}=-1.051V$

Therefore, the value of $E^0$  of anode half-cell is, -1.051 V