Write the half reaction and balanced equation for:

Question

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.

Answer 1

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

Answer 2

In cell notation, the species are written in the order of their presence in the cell from left to right. The anode is always written on the left side and represents the half-reaction that occurs at the electrode where oxidation takes place. Similarly, the cathode is written on the right side and represents the half-reaction that occurs at the electrode where reduction takes place.

In the given reaction, the oxidation half-reaction is:

H2(g) -> 2H+ + 2e-

And the reduction half-reaction is:

Cl2(g) + 2e- -> 2Cl-

Now, when writing the cell notation, we start with the anode species on the left side, which is H2(g). Next, the aqueous species are written, and since H+ is also present in the anode compartment, it is written next. Then we use the double line || to separate the anode from the cathode.

On the right side, we start with the cathode species, which is Cl-. Next, we write Cl2(g) since it participates in the reduction half-reaction. Finally, we end with the Pt(s) electrode on the right side.

So the correct cell notation would be:

Pt(s)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s)

This notation correctly represents the anode (oxidation) and cathode (reduction) sides of the cell.

Answer 3

In the given redox reaction, the half-reaction for the oxidation of hydrogen (H2) to form hydrogen ions (H+) is correct:

H2(g) -> 2H+ + 2e-

However, the half-reaction for the reduction of chlorine (Cl2) to form chloride ions (Cl-) needs to be adjusted. The correct half-reaction for the reduction of chlorine is:

Cl2(g) + 2e- -> 2Cl-

To balance the charges, two electrons (2e-) are needed on the reactant side.

Now, let's move on to the cell notation. Your understanding of the general format of cell notation is correct, which is anode | anode species | anode charge || cathode species | cathode charge | cathode. However, the placement of the species within the cell notation depends on their respective positions in the electrochemical cell.

In the given reaction, hydrogen (H2) is oxidized at the anode, whereas chlorine (Cl2) is reduced at the cathode. Therefore, the cell notation would look like this:

Pt(s)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s)

The anode is placed before the cathode in the cell notation because the oxidation reaction (anode reaction) occurs first, followed by the reduction reaction (cathode reaction).

So, the answer in the book "Pt(s)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s)" is the correct cell notation for your given redox reaction.