For the following battery:

Cd(s) | CdCl2(aq) || Cl–(aq) | Cl2(l) | C(s)

Write the reduction half reaction for the C(s) electrode.

Ah, the mighty C(s) electrode! It's always at the center of attention, isn't it? Now, let's see what it's up to in this battery party.

Well, my friend, the reduction half-reaction for the C(s) electrode goes a little something like this:

C(s) + 2Cl^-(aq) → Cl2(l) + 2e-

Yep, C(s) is getting all cozy with a couple of Cl^- ions and turning them into lovely Cl2 molecules while generously donating a couple of electrons. It's like playing matchmaker for chemicals!

The reduction half reaction for the C(s) electrode can be written as follows:

C(s) + 2e- → 2Cl-(aq)

To determine the reduction half reaction for the C(s) electrode in the given battery, you need to know the standard reduction potential values for the different reactions involved.

1. Start by identifying the C(s) electrode and determine its half reaction using the standard reduction potential table.
- In this case, the C(s) electrode represents a solid carbon electrode.

2. Look for the half reaction involving carbon in the reduction half cell. In the standard reduction potential table, the closest available reaction involving carbon is shown below:
C(graphite) + 4e- -> C(s)

3. However, this reaction involves carbon in the form of graphite, not a solid carbon electrode. Therefore, we need to adjust the reaction to match the given electrode:
C(graphite) + 4e- -> C(s)

So, the reduction half reaction for the C(s) electrode is:
C(graphite) + 4e- -> C(s)

Please note that the standard reduction potential values determine the direction of the cell reaction, and if the reduction half cell involves a solid electrode, it's often represented as C(s) for simplicity.

Cl2 + 2e ==> 2Cl^-