I need help figuring out the standard reduction potential?

Question

This is a series of electrochemical half-reactions and found the half-cell reduction potential for all but one, ClO2(aq) → ClO-1(aq). When you couple this half-cell to a MnO4-1(aq) → Mn+2(aq) half-cell as the cathode (Ered = +1.362V), you find that the measured cell voltage is +1.218V.

A) After the experiment was done you noticed that the permanganate half-cell was connected as the anode in the voltaic cell. Now, write a balanced chemical equation for each half cell and the net overall reaction when the permanganate half-cell is the anode.

**For this the answer I got for the balance equation was: 2 Mn²⁺+ 3 H₂O + 5ClO₂ --> 2 MnO₄^- + 6 H⁺ + 5ClO^-

B) What is the standard reduction potential for your ClO2(aq) → ClO-1(aq) half-cell if permanganate was the anode?

**I got stuck on this part and I need help figuring out writing the standard reduction ClO2(aq) → ClO-1 with permanganate anode. I know that answer for ClO2(aq) → ClO-1(aq) half-cell is -1.144

Answer 1

A)

If the MnO₄^- ==> Mn²⁺ was actually the anode and not the cathode, then we can write the oxidation reaction (anode) as follows:

Mn²⁺ + 4H₂O ==> MnO₄^- + 8H⁺ + 5e- Eº = -1.362 V (change sign since reaction is reversed)

We can write the reduction (cathode) half reaction as follows:

ClO₂ + 2H⁺ + 3e- ==> ClO^- + H₂O Eº = ?

Overall reaction:

3Mn²⁺ + 7H2O + 5ClO₂ ==> 3MnO₄^- + 14H⁺ + 5ClO^-

B).

Since Eºcell = Eºcathode - Eºanode, we can determine Eº for ClO₂ ==> ClO^- as follows:

1.218 = Eºcathode - (-1.362)

1.218 = Eºcathode + 1.362

Eºcathode = 1.218 - 1.362

Eºcathode = -0.144 V = standard reduction potential for ClO₂ ==> ClO^-

So, my answer is different than what you report by 1.00, so I'm wondering why that is unless you meant -0.144 instead of 1.144? Maybe someone else will chime in on this.