Nickel and aluminum electrodes are used to build a galvanic cell.
A) What is the theoretical cell potential assuming standard conditions?
B)Type the shorthand notation for this cell. Do not include concentrations.
For example, in shorthand notation your answer might look like Cu|Cu^+||Pd^(2+)|Pd.
2 Answers
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=So i’m going to make the assumption that the cell must be setup such that the reaction will be spontaneous.
The first thing you need to do is find a table of standard reduction potentials and look up the reactions involving iron and alumminum, as the name of the table implies, both of these will be written as reduction reactions. The reactions are below
Al3+ + 3e- —> Al E(0) = -1.66
Ni2+ + 2e- —> Ni E(0) = -0.28
so just like Gibb’s free energy, there are certian limits on the E(0) of the cell that determine the spontaneity of the reaction. Moreover, if E(0) which is the net voltage or potential is greater than 0 the reaction is nonspontaneous, if it is less than 0 it is spontaneous, and if it is 0, then the reaction is at equilibrium. So it should make sense that we want the half reaction (that is the reactions with alumminum and nickel) that has a more negative value to remain the reduction and the one which is less negative to become an oxidation. To switch a reaction to an oxidation, you reverse it and change the sign on its E(0), doing this, our cell now becomes:
Al3+ + 3e- —> Al E(0) = -1.66 REDUCTION
Ni —> Ni2+ + 2e- E(0) = +0.28 OXIDATION
but as you can see, the alumminum gains three elctrons while nickel only loses two, so we have to adjust for this by multiplying the reactions until we have a common number of electrons. The lowest common multiple of 3 and 2 is obviously 6, so we’ll multiply the reduction by 2 and the oxidation by 3, yielding:
2 Al3+ + 6e- —> 2 Al E(0) = -1.66
3 Ni —-> 3 Ni2+ + 6e- E(0) = +0.28
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2 Al3+ + 3Ni —> 2 Al + 3Ni2+ E(cell) = 1.38
note that when you multiply the reactions, the potential remains unchanged; therefore, the theoretical potential of the cell is -1.38.
The shorthand notation is written like this
Al3+/Al//Ni/Ni2+ where the / represents a phase change (vertical line) and // represents two different half reactions. When writing these reaction diagrams the anode (oxidation) is always written before the cathode (reduction).
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Technically this answer is correct. But MasteringChemistry is a stupid foul program, and it must be typed in EXACTLY like this for credit:
Al|Al^3+||Ni^2+|Ni
Source(s): MasteringChemistry
