## Carbon Steel as an Interstitial Alloy and How its Chemical Structure Contributes to Elasticity of Collisions

### Background

Back when our high school physics class was learning about elastic and inelastic collisions with the conservation of momentum, I noticed that a disproportionate number of the practice problems we were assigned that had to do with elastic collisions involved steel. Obviously, when we think of steel we think of its “hardness.” It is used in a lot of heavy-duty construction, after all.

It just so happened that at precisely this time, we were learning about alloys in chemistry, and I remembered specifically my teacher talking about steel being an alloy of iron and carbon. Thus, I had a little flicker of inspiration and decided to connect the two and make a little presentation for my physics class. An added bonus? A little bit of extra credit. Continue reading Carbon Steel as an Interstitial Alloy and How its Chemical Structure Contributes to Elasticity of Collisions

## Salt Bridges in Galvanic Cells

As we may already know, galvanic cells are simply experimental setups that take advantage of the electron transfer during a redox reaction to do electrical work. It may be quite easy to grasp that idea that electrons flow from the anode to the cathode, but the salt bridge is a little less easy to understand.

For any of you out there that are just looking for an easy way to remember that you need a salt bridge for any galvanic cell to function properly, something you can remember is that the salt bridge simply completes the circuit. It provides that “return route” for electrical charge to travel back into the anode.

Chemistry textbooks often say that a salt bridge is necessary to “balance out the charges in the two half-cells” but what does this really mean? Let us look at the basic set-up of a galvanic cell again.

As the redox reaction occurs, electrons migrate from the anode to the cathode. In our specific case with $Zn_{(s)}$ as our anode, as the reaction carries on, more and more $Zn^{2+}_{(aq)}$ is formed. The net result is that the left half-cell becomes more and more positive. You can think of this in two ways:

1. Electrons are being taken out of the left side, and so the left side becomes more positive by virtue of subtracting negative entities.
2. $Zn^{2+}_{(aq)}$ is being formed on the left side and because they are cations, it becomes more positive.

Similarly, but in the exact opposite way, on the right half-cell $Cu^{2+}_{(aq)}$ is reduced to $Cu_{(s)}$. The net result is that the right side becomes more and more negative. Again, you can think of this in two ways:

1. Electrons are being added to the right side, and because we are adding negative-charged particles the right half-cell must become more negative.
2. $Cu^{2+}_{(aq)}$ (positive) is becoming $Cu_{(s)}$ (neutral) and so we are losing positive charge on the right, and thus becoming more negative.

By virtue of the left half-cell becoming positive and the right half-cell becoming negative, the electrons would be less inclined to move from the anode to the cathode. After all, why would a negative particle spontaneously move from something that is positive to something that is negative?

This is precisely where the salt bridge kicks in. The salt bridge contains both cations and anions (in our case $KCl$) that can enter either side as the reaction progress continues. Because the left half-cell becomes more positive, anions (negative) from the salt bridge will migrate there. The opposite occurs in the right half-cell. Given $KCl$, the $Cl^{-}$ will be attracted to the positive left half-cell and the $K^{+}$ will be attracted to the negative right half-cell.

Understanding this fact will help you grasp the textbook wording of “balancing out the charges in the two half-cells.” It helps keep the left side negative and the right side positive… for the happiest of balances!