<p>For #3, can somebody PLEASE explain to me how electrochem/salt bridges work?? I don't get ANYTHING!!!!!!11897reuhfijhgjalfd....</p>
<p>like how the helk do I know that the electron flow is to from right to left??</p>
<p>Generally, it is always right to left in galvanic cells. To determine whether an ion is an anode, check its reduction half-rxn on the other side of the periodic table, and see if it is the least positive - most negative, in comparison to the other ion. Flip that charge so it turns from negative to positive, because an anode oxidizes, and that’s how you determine if it is an anode. </p>
<p>The reverse goes for an electrochemical cell, in determining the anode, because it requires an external energy source. Hope this helps :D</p>
<p>Serafina, how do I know if that is a galvanic cell?</p>
<p>A galvanic cell does not have an external battery source. You correctly identified that FRQ question as being electrochemical, since it distinctly said “external battery source.”</p>
<p>so you are saying the more positive standard reduction potential (we determine this from the chart they give us right?) is the anode?</p>
<p>so since O2+4H+4e is higher up on the chart, that’s the anode?</p>
<p>WAIT… so if the diagram is an electrolytic cell, then shouldnt the flow be from cathode to anode? and since Cu^2+ + 2e- has a lower standard reduction potential, shouldnt the flow be from left to right?.. i dont get it!!</p>
<p>That’s the beauty of electrochemistry, it’s almost impossible to get!</p>
<p>[SparkNotes:</a> Electrolytic Cells: Electrolysis](<a href=“http://www.sparknotes.com/chemistry/electrochemistry/electrolytic/section1.html]SparkNotes:”>http://www.sparknotes.com/chemistry/electrochemistry/electrolytic/section1.html)</p>
<p>Electrolysis is essentially the entire reverse of galvanic cell batteries.</p>
<p>I appreciate your help… but I still need a clearer explanation of determining which is the cathode/anode and determining if a cell is galvanic vs. electrolytic. </p>
<p>Any help is appreciated.</p>
<p>I’ll do this, since I have to beat electrochemistry concepts in my head as well.</p>
<p>Anodes are pure metals that lose electrons through oxidization.</p>
<p>Zn -> Zn2+ + 2e-</p>
<p>Cathodes are metal ions that gain electrons from the anode half-cell
Cu2+ + 2e- -> Cu</p>
<p>That is more clearly defined in the diagram that i linked to above. For a galvanic cell (not linked to outside energy source), determine whether an ion is an anode, check its reduction half-rxn on the other side of the periodic table, and see if it is the least positive - most negative, in comparison to the other ion. Flip that charge so it turns from negative to positive, because an anode oxidizes, and that’s how you determine if it is an anode. </p>
<p>The anode always flows to the cathode, no matter if it is an electrolysis or a galvanic cell, since cathodes are electron-deficient and starve for the excess electrons given off from anodes.</p>
<p>The reverse is for an electrochemical cell, since an outside energy source basically reverses the entire battery’s reactions and flow of electrons. The position of the anode cathode get flipped, so it is cathode anode, and the most positive reduction half rxn of the two metal ions becomes negative, becoming the anode. That’s the explanation for why it flows right to left. </p>
<p>When calculating Ecell = Eoxidization - Ereduction, it’s basically :</p>
<p>Ecell = Charge of anode - charge of cathode. So it would be H2O… - Cu2+, in this case, since H2O is the anode on the right side o the electrolysis cell, and Cu2+ is the cathode on the left side of the electrolysis cell. Again, anodes flow to cathodes, since the loss of electrons from the anodes flow to the electron-starving cathodes.</p>
<p>okay this is how i would do it. </p>
<p>The basic: cathode is reduction, anode is oxidation. reduction has e- on the left side of equation, oxidation has e- on the right. galvanic is spontaneous = positive E.</p>
<p>Electrons always flow from anode to cathode! no matter galvanic or electrolytic. Reference your textbook for the full explanation about adding mass, increasing concentration of ion. </p>
<p>So you know Cu is being reduced (from the equation next to the picture. Election on the left side of arrow). This must mean that H2O is oxidized. So, this means that you have to flip the half equation going to H2O given in the box. The E value then becomes the opposite, -1.23. </p>
<p>emf = cathode + (- anode). Since cathode is 0.034 (reduction), anode is 1.23. When you add the two up, you get a negative number.</p>
<p>MEMORIZE- negative E means non spontaneous = electrolytic. Or you could try the short cut: if there’s a power supply, it’s non-spontaneous.</p>
<p>I support velleity’s response. We basically gave two different ways to explain this process, so I hope it works out better?</p>
<p>velleity. Thank you. That helped a lot. </p>
<p>One more question, though. Yes it does have an electron on the left in the equation next to the picture, but if you look at the box at the bottom, both Cu and O2 have electrons on the left. So if the V is more positive, is it anode?</p>
<p>In that case, yeah, you must compare the E (Volt) values. Flip whichever to make the total negative. The one you flip is the anode. </p>
<p>For a spontaneous reaction, the most positive reduction is the “actual” reduction, or cathode. The opposite would go for non-spontaneous. But somehow I don’t recall ever having to do that.</p>
<p>Whoa whoa whoa… wait… </p>
<p>How do I know I have to make the cell negative?? I don’t know that it’s electrolytic yet, do I?</p>
<p>What i mean is, I can only know that the cell is electrolytic AFTER i see that the total V is negative. What you told me is, make the V negative since I know that the cell is electrolytic, but I don’t.</p>
<p>That will NEVER be the case. They will tell you one way or another the spontaneity of the reaction. </p>
<p>Option one: draw you a power supply
Option two: tell you one of the two half reaction
Option three: include the word in the question</p>
<h1>3 tells you option 1 and 2.</h1>
<p>The cell is electrolytic when there is an external power source, powerbomb. That’s the criteria. Everything else comes after that.</p>
<p>oooooh alright. THANKS!</p>