I. Your job first is to make two electrochemical cells with the maximum voltage
ID: 1052023 • Letter: I
Question
I. Your job first is to make two electrochemical cells with the maximum voltage (E) possible with the equipment you are given. Choosing the substances to go in the half-cells: It would make sense to think about this ahead of time and use some theory to choose a most appropriate system. You want to find the two half-reactions that predict the largest cell voltage; if concentrations are all 1M, the highest voltage would be obtained using best oxidizing agent and the best reducing agent. However, to use the predictions, your cell should, at the beginning, contain some of everything in the overall cell reaction equation. What I mean by this is illustrated with an example: suppose you are considering using Cd and Cd2 in your cell because the Cda- +2 e- Cd half reaction has a favorable voltage, but you have no Cd available. Because you have no Cd available, you cannot build your cell with it, so eliminate that half-reaction from your list of possibilities. Does concentration matter? After you pick 3-4 possible cells to make, you should use the Nernst equation to consider the effect of concentrations available: you may also dilute any of the provided solutions if doing so gives a larger voltage. In deciding whether to do this, consider the algebraic form of the Nernst equation and decide if diluting reactant or diluting product would have any affect on the voltage and if the effects are the one you want (maximizing voltage). If it helps get a larger voltage, you may dilute your solutions (not the stock bottle, of course) or add acid. Once you decide which two cells are your top candidates, you should make them and measure their voltages. Your report should explain how you used chemical principles to decide on those two possibilities and should calculate a percent error. After you make the two cells, show your instructor what voltage they produce. Available equipment. porous cups (see below for how these are used) Electrodes available zinc Znls iron F nickel nils copper Cat Carbon (may be regarded as inert) uminum AL aluminum Solutions available: Do not use any more than the listed amounts in a cell. These ar columns for any particular reason-they may be used in any combination you wish 5mL of 0.10 M CuS04 5 mL of 0.1 M FeCI3 5 mL of 0.1 M NiSO e not listed in 5 mL of 0.10 M ZnSO4 5 mL of 0.1 M FesO4 49Explanation / Answer
Looking at the standard reduction potential chart,
the more positive the electrode potential of a half-reaction is, stronger is the oxidizing agent.
the more negative is the standard reduction potential of a half-reaction is, the stronger is the reducing agent.
Thus highest potential cells would be,
Al | Al3+ || Fe | Fe3+
the electrode potential for the cell (Eo) = -0.037 - (-1.66) = 1.623 V
Al | Al3+ || Cu2+ | Cu
the electrode potential for the cell (Eo) = 0.337 - (-1.66) = 1.997 V
This would be the cell with the highest electrode potential of all.
Nernst equation,
E = Eo - 0.0592/n logK
So as the concentration of product increases, the K value increases, the overall logK increases and this would reduce the E value for the electrochemical cell.