You are right that the anode is on the left and the cathode on the right. But the cell notation is written as they appear from left to right as you go from the anode, across the salt bridge, and to the cathode.
In the cell you will have Pt on the left and Pt on the right. On the left you have H2 bubbled INTO H^+. On the right you have Cl2 bubbled INTO Cl^-. Therefore, from the left electrode across the salt bridge you have Pt, H2, Cl^- Cl2,Pt.
You can read more here.
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Cell-Notation-and-Conventions-995.html
Write the half reaction and balanced equation for:
Cl2(g) + H2(g) -> HCl(aq)
so I got:
2e- + Cl2(g) -> 2Cl-
H2(g) -> 2H+ + 2e-
so Hydrogen is oxidized and Chlorine is reduced
so for cell notation I put:
Pt(s)|H2(g)|H+(aq)||Cl2(g)|Cl-(aq)|Pt(s)
but the answer in the book is:
Pt(s)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s)
I was under the assumption that cell notation is written with anode, then cathode. Then, the elements are written in the order of reactants, then products.
Why is Cl- before Cl2, if Cl2 is the reactant that gets reduced?
Thanks.
1 answer