Consider a galvanic cell with a beaker of Chromium(III) nitrate Cr(NO3)3 and a beaker of Copper(II) nitrate Cu(NO3)2. The Chromium(III) nitrate beaker contains a strip of chromium, and the Copper(II) nitrate cell contains a strip of copper. A wire runs between the strips. The reaction that occurs is as follows:
2Cr(s) + 3Cu2+(aq) → 2Cr3+(aq) + 3Cu(s)
In three to five sentences, list which electrode is the anode and which is the cathode and the half reactions that occur at each electrode.
The Chromium strip is the anode and the Copper strip is the cathode. At the anode, the half reaction is 2Cr(s) → 2Cr3+(aq) + 6e-. At the cathode, the half reaction is 3Cu2+(aq) + 6e- → 3Cu(s).
Well, this galvanic cell is quite the fancy setup! Now, let's get to the heart of the matter - the electrodes. In this case, the chromium strip is the anode (pronounced "a-node"), which is the negative electrode. It's a little like the “no” in "anode" - always saying "no" to positive charges. The half reaction that occurs at the anode is 2Cr(s) → 2Cr3+(aq) + 6e-. On the other hand, the copper strip is the cathode (pronounced "cat-hode"), which is the positive electrode. It's like the "cat" of the cell, attracting positive charge like a magnet. The half reaction that occurs at the cathode is 3Cu2+(aq) + 6e- → 3Cu(s). It's like the copper ions saying goodbye to their positive charge and becoming solid copper. Voila!
In this galvanic cell, the anode is the electrode where oxidation takes place, and the cathode is the electrode where reduction takes place. In this case, the anode is the strip of chromium, and the cathode is the strip of copper.
At the anode (Chromium electrode), oxidation occurs, so the half-reaction is:
2Cr(s) -> 2Cr3+(aq) + 6e-
At the cathode (Copper electrode), reduction occurs, so the half-reaction is:
3Cu2+(aq) + 6e- -> 3Cu(s)