An electrochemical cell is made from two half cells. The first contains 1M hydrochloric acid above 1 atm Cl2gas (w/ a platinum electrode). The other half cell contains a solution of H2O2 above 1 atm of O2 (also with a platinum electrode). A spontaneous reaction occurs.

a. Write the balanced redox reaction

(having trouble figuring this out)

What about

Cl2 + H2O2 ==> 2Cl^- + O2 + 2H^+
I don't know if this is the spontaneous direction or the reverse but you can look up the Eo potentials and decide which direction to write it.

Thank you!

To write the balanced redox reaction for the given electrochemical cell, you need to identify the oxidation and reduction half-reactions occurring in the cell.

In the first half-cell:
1M hydrochloric acid (HCl) is present with 1 atm of chlorine gas (Cl2) and a platinum electrode.
The possible half-reactions are:
1) Cl2(g) → 2Cl^-
2) Cl2(g) + 2e^- → 2Cl^-

In the second half-cell:
A solution of hydrogen peroxide (H2O2) is present with 1 atm of oxygen gas (O2) and a platinum electrode.
The possible half-reactions are:
3) 2H2O2(aq) → O2(g) + 2H2O(l)
4) O2(g) + 4H^+ + 4e^- → 2H2O(l)

Now, to determine the actual redox reaction, you need to match the number of electrons transferred in each half-reaction.

Comparing half-reactions (1) and (4), they both involve the transfer of 2 electrons. Thus, we can conclude that half-reaction (1) with Cl2 is oxidized, while half-reaction (4) with O2 is reduced.

Balancing the redox reaction:
Multiply half-reaction (1) by 2 to balance the electrons.
2Cl2(g) → 4Cl^-

Now, add the balanced half-reactions:
2Cl2(g) + 4H^+ + 4e^- → 4Cl^- + 2H2O(l)
2H2O2(aq) → O2(g) + 2H2O(l)

The final balanced redox reaction is:
2Cl2(g) + 4H^+ + 4e^- + 2H2O2(aq) → 4Cl^- + O2(g) + 2H2O(l)

This is the balanced redox reaction for the given electrochemical cell.