I observed that the copper penny seemed to have stayed the same while the scraped sides of the copper coated zinc penny deteriorated and were black. I was told that a single replacement reaction occurred. Explanations?
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Zinc reacts with mineral acids. Copper doesn't (except oxidizing acids). Check out the activity series of metals. Those above hydrogen can replace it from acid and form their respective salt.
I assume you removed enough of the copper to expose the zinc core. The zinc will react with HCl, but copper won't, at least that is what we are supposed to believe. If you leave the copper penny in the HCl long enough the penny will disappear leaving a green solution. But that requires a while to occur. In the first coupe of hours or even days, the copper will appear not to react, which is what we would predict from the activity series and the position of Cu compared to H. I've taught chemistry for nearly 40 years and I've dissolved the zinc out of the middle of countless pennies. The removal of the zinc takes several days, but if you leave it in the HCl too long the copper will dissolve as well, which is counter to what the activity series tells us. The reason for the copper reacting isn't the reduction of H+ to H2 gas as is the reaction for zinc metal. Instead, there is a thin coating of CuO on the copper penny which reacts with HCl to make Cu2+ ions in solution. Cu2+ will react with Cu metal to make Cu+ which reats with Cl- to make insoluble CuCl. CuCl reacts with oxygen to give CuO and CuCl2(aq). Gradually, the copper metal is converted to Cu2+. In the presence of concentrated Cl- ions we find CuCl4^2- which is yellow. Since there will also be some [Cu(H2O)6]^2+ which is blue, the combination of yellow and blue will appear green, which is what we observe.
