Electrochemistry
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Hi, I have a question on electrochemistry that I am practising.
It is about an electrolytic cell.
Electrode X is connected to the positive cell of the battery.
Electrode Y is connected to the negative cell of the battery.
Electrolyte used is 1 mol dm-3 KI(aq).
Qn1: What reactions will occur at electrode X and electrode Y?
Qn2: As the reaction proceeds, a white coating is formed on one of the electrodes together with a brown solution. Suggest what has happened to give the above observations.Thank you :)
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For ElectroChem, always remember the image of a "Red Cat riding An Ox."
Electrons from the battery (which travels from negative to positive cell of battery) travel to the electrode Y, and are used to reduce protons from water to hydrogen gas, since K+ is too stable (or K is too reactive) to be reduced in an aqueous solution.
At X, bromide ions are oxidized to bromine molecules. The electrons released from the oxidation of bromide ions travel to the positive cell of the battery. Hence, the brown solution.
The electrode on which the "white coating" is observed, is constructed of a metal which will form a precipitate with hydroxide ions (which are produced as protons are removed from water molecules at the cathode), eg. zinc electrode reacts to form white zinc hydroxide ppt which will dissolve in excess hydroxide ions due to formation of a soluble tetrahydroxy complex ion.
Finally, in electrochem questions, always remember to quote relevant standard reduction/oxidation/electrode potentials from the Data Booklet.
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Since the electrolyte is aqueous KI (potassium iodide), shouldn't it be iodide ions which are oxidised to iodine molecules at electrode X?
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Yes, you're right. An oversight there. Iodide ions oxidized to iodine molecules, hence the brown solution.
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Sorry I forgot to add that BOTH the electrodes are made of copper.
I am alittle confused with which equations to use.
At the cathode, if you say that water is reduced to H2.
I would use the equation:
2H20 +2e <--> H2 + 2OH- E= -0.83V
Is this right?
Still, I am not sure what the white coating is, but I am guessing along the line of Cu (I) iodide? hahaha. -
Correct on both counts.
Regarding copper(I) iodide, visit :
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Hmm..but I dont quite understand the equations involved.
Anode: Cu2+ + 2e<-> Cu E =+0.34V
Cathode: 2H20 +2e <--> H2 + 2OH- E= -0.83V
Overall E(cell) = -1.17V (<0, how does the cell works?)
On top of that, how is Copper(I) iodide formed in the process, when at the anode there is only Cu2+? and the brown solution?
What other processes are occuring? -
At the anode, Cu metal is oxidized to Cu2+ ions.
The Cu2+ ions then bond ionically with the iodide ions to form copper(II) iodide.
Cu2+ + 2I− → CuI2
Being unstable, the copper(II) iodide decomposes to iodine and insoluble copper(I) iodide, releasing I2.
2 CuI2 → 2 CuI + I2
See :
http://en.wikipedia.org/wiki/Copper(I)_iodide
Reduction Potential for reduction of H2O to H2 gas and OH- ions is -0.83V
Oxidation Potential for oxidation of Cu to Cu2+ is -0.34V
Standard Electromotive Force (EMF) of Cell is (-0.34)+(-0.83) = -1.17V
(Alternatively, use the formula E-Standard of Cell = Reduction Potential at Cathode - Reduction Potential at Anode. (-0.83)-(+0.34)=-1.17V)
You're wondering how the current can flow in this cell, given that the overall redox is neither spontaneous nor feasible under standard conditions.
No worries. You forgot that this is an ELECTROLYTIC cell, not a GALVANNIC aka VOLTAIC cell.
In an Galvanic aka Voltaic cell, the natural redox reactions that occur at the anode and cathode, drive the current.
In an Electrolytic cell, the electrons are forced to flow against their natural redox tendencies, by use of a battery (or external current supply).
Hence, the E-standard (commonly mispronounced as e-naught; the 'o' superscript symbol isn't a zero; it's a zero with a horizontal slash, meaning 'under standard conditions' of 1atm for gases, and 1 mol/dm3 for all solutions; note that the term 'standard conditions' by itself does not automatically specify temperature (another common misconception); the values in the data booklet are for 25 deg C *and* under standard conditions).
Hence, for an ELECTROLYTIC cell, the E-standard or standard ElectroMotive Force (EMF) of the Cell, need not be positive for the current to flow - the battery is the driving force behind the current flow.
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Oh yes.
I forgot that this is an electrolytic cell hence there is no need to bother about the E (standard).
Thanks.
Didn't realize that Cu(II) iodide is unstable, always thought Cu2+ is a stable ion for copper. -
I have another question:- :)
Will the cell e.m.f. be affected when the size of electrode changes?
Say for this question above, the size of the copper electrode increases?
My initial guess is no change in em.f because erm.....because, the position of equilibrium of the Cu/Cu2+ did not change?
Thanks. -
Correct. The only factors that will change e.m.f. is molarity of solutions (Data Booklet values are for 1 mol/dm3), pressures of gases (if gaseous reactants are involved), temperature (Data Booklet values are for 25 deg C), or adding other compounds that would react to remove one or more of reactants/ions, causing a shift in equilibrium, etc.
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may i ask..
electrolysis is carried out with the electrodes, magnesium and zinc.
of course, electrons will flow from magnesium to zinc.
is it that current flows from zinc to magnesium because electrons always flow the opposite direction to current flow?
and if so, doesn't that make zinc the anode instead of the cathode?
the answer said that zinc is the cathode which makes me really confused. please help!!
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Originally posted by YTH:
may i ask..
electrolysis is carried out with the electrodes, magnesium and zinc.
of course, electrons will flow from magnesium to zinc.
is it that current flows from zinc to magnesium because electrons always flow the opposite direction to current flow?
and if so, doesn't that make zinc the anode instead of the cathode?
the answer said that zinc is the cathode which makes me really confused. please help!!
Always visualize a "Red Cat riding An Ox". Reduction occurs at the Cathode. At the Anode, Oxidation occurs.
If this is an electrolytic cell (as opposed to an electric cell aka Galvanic cell aka Voltaic cell), then which electrode is the cathode and which is the anode, depends entirely on how you place your battery. Remember electrons flow from negative terminal to positive terminal.
If this is an electric cell aka Galvanic cell aka Voltaic cell, then (you should be able to think out the following step-by-step process on your own in the exams), the following applies :
All metals (ie. atomic form) want to oxidize themselves into stable cations.
Magnesium, being more reactive than zinc, would bully zinc into accepting it's valence electrons as it happily throwns them off, and being oxidized to Mg2+ ions.
Hence, magnesium is the Anode (Oxidation).
This forces zinc to be the Cathode (Reduction). Since zinc metal is already in its reduced state, it cannot be reduced further. But being the cathode, something MUST be reduced here (Red Cat). So the cations from solution (eg. H+, Cu2+ etc), or water itself, will undergo reduction at the zinc cathode.
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thank you so much! i can understand this much better now!
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Originally posted by YTH:
thank you so much! i can understand this much better now!
You're totally welcome, YTH!
(All you Sec 4 graduating folks, if you're going JC next year and would like 'A' level Chem tuition, you know where to contact me...
Pssst. It's really true that how you see and approach JC Chemistry, really makes a BIG difference whether you'll enjoy JC Chemistry or not. I help the students to see that Chemistry's really actually enjoyable, where JCs make it (falsely) seem a boring torture.)