Redox and electrochemical cells Module 5 Flashcards
What’s reduction?
Gain of electrons
What’s oxidation?
Loss of electrons
Which is the oxidising agent?
The one which accepts electrons and has been reduced
Which is the reducing agent?
The one which donates electrons so has been oxidised
How would you separate redox reactions into half equations?
Split it into a oxidation half equation (electrons are lost so on product side)
and a reduction half equation (electrons are gained so are on reactant side)
How can you use redox half equations to form a full redox equation?
Add them together and cross the electrons out
Take into account that if one has more electrons than the other will have to multiply up so they are even
The amount of atoms also need to balance on each so can add water and H+ ions to this, and then at the end can add e-‘s to balance the charge
What does an electrochemical cell contain?
2 different metals dipped in salt solutions of their own ions, connected with a wire and salt bridge, and voltmeter in the circuit
Why is platinum used in a electrochemical cell sometimes?
It’s inert and conducts electricity
What happens in a Zn and Cu half cell?
Zinc loses electrons more easily than copper, so in it’s half cell Zinc (from the electrode) is oxidised to form Zn 2+ ion, this releases electrons into the circuit
In Cu half of the cell the same amount of electrons are required to reduce Cu2+, using the electrons from thee external circuit
So electrons flow from Zn to Cu
What’s a salt bridge and what does it do?
Filer paper soaked in KNO3
Allows ions to flow through and balance the charge
What do you if solution contains 2 different (aq) ions?
Use platinum electrode
What do you do if one of parts of half equation is a non metal?
Use platinum electrode
Have (aq) solution of ions (the same)
Bubble gas over platinum electrode
The reactions that occur at each electrode are? And what is the standard notation?
Reversible
Always put e- on the reactant side
What do you use to know how easy a metal is to oxidise?
E⦵ value, the more negative it is the easy it is to oxidise
So the more negative E⦵value reaction goes backwards
And the more positive E⦵ goes forwards
How do you calculate the E⦵ value of a cell?
E⦵ = E⦵ of more positive side - E⦵ of more negative side
So E⦵ will always be positive
Standard electrode potential definition?
E⦵ of a half cell is the voltage measured under standard conditions when the half cell is connected to a standard hydrogen electrode (0.00v)
What are the standard conditions in terms of electrochemical cells?
Any solutions have a concentration of 1.00 mol dm^-3
Temperature must be 25 degrees (298 kelvin)
Pressure must be 100kPa
As disrupts the equillibriums
Do “more reactive” or “less reactive” metals have a more negative standard electrode value?
More reactive
How to predict whether or not a chemical reaction will happen or not?
Write down both half equations with electrons on the standard left size
Look at the E⦵ value, the more positive one the reaction goes forwards (left to right)
The more negative one goes backwards (right to left)
Combine the 2 half equations remembering to multiply if uneven amount of electrons
Check to see if the combination of half equations matches what has been asked in the question
If it does then it’s feasible, if doesn’t match it’s not
Reasons why a E⦵ prediction may be wrong?
The conditions are not standard
The rate of reaction is so slow reaction doesn’t appear to be happening
Very high activation energy
Explain changes that may occur in a E⦵ cell prediction if conditions are not standard?
Eg for the equillibirum Zn(s) + Cu(2+) =(reversible) Zn(2+) + Cu
If you increase the concentration of Zn(2+) equillibrium shifts left, reducing the ease of electron loss, so the electrode potential of Zn/Zn(2+) becomes less negative and the whole cell potential will be lower
If you increase the concentration of Cu(2+), the equillibrium will shift right, increasing the ease of electron gain, the electrode potential of Cu(2+)Cu becomes more positive and the whole cell potential becomes higher
What does a more negative E⦵ value mean?
Wants to lose electrons more to form a positive ion
to be oxidised
Instead of working out cell potential, in batteries you work out?
Cell voltage
Describe how fuel cells generate electricity?
At the anode the platinum catalyst splits the H2 into protons and electrons (H2 = 2e- +2H(+))
The polymer electrolyte membrane only allows the H+ across and forces the e- to travel around the whole circuit to get to the cathode
An electric current is created in the circuit, which is used to power something
At the Cathode, O2 combines with the H+ from the anode and the e- from the circuit, to form H2O which is the only waste product
Advantages of electrochemical cells?
They are more efficient at producing energy than combustible engines
Produce a lot less polluition
Negatives of electrochemical cells?
The production of the cells involves the use of toxic chemicals which needs to be disposed of once it’s finished it’s lifespan
They are very flammable
What are the 2 half equations for the redox titration of Acidified Potassium manganate oxidising Fe(2+) to Fe(3+) and the colour change and what does this titration tell us?
MnO4(-) + 8H(+) + 5e- = Mn(2+) + 4H2O
5Fe(2+) = 5Fe(3+) + 5 e-
So equals MnO4(-) + 8H(+) + 5Fe(2+) = Mn(2+) +4H2O + 5Fe(3+)
Once the reaction has finished will see a colour change from purple to colourless as MnO4(-) is pink and {Mn(H2O)6}(2+) is colourless
How much Oxidising agent we need to reduce the species
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What are the 2 half equations for Potassium dichromate solution K2Cr2O7 oxidising Zn to Zn(2+) and the colour change and what does this titration tell us?
Cr2O7(2-) + 14H+ + 6e- = 2Cr(3+) + 7H2O
3Zn = 3Zn(2+) +6e-
So equals Cr2O7(2-) + 14H+ + 3Zn = 2Cr(3+) + 7H2O + 3Zn(2+)
Colour change when reaction is finished will be orange to green as Cr2O7(2-) is orange and {Cr(H2O)6}(3+) looks green
How much Oxidising agent we need to reduce the species
All the steps and equations for a sodium thiosulfate titration to find the concentration of an oxidising agent?
Use a certain volume Potassium Iodate solution to oxidise an excess of KI
IO3(-) + 5I(-) + 6H+ = 3I2 + 3H2O
Find out how many moles of iodine have been produced, by titrating the resulting solution with a known concentration of Sodium Thiosulfate (Na2S2O3), when the solution goes pale yellow showing it’s near to the end point, add starch solution, which will show that Iodine is still present by being blue/black, will know reaction is finished when blue/black dissapears
I2 + 2S2O3(2-) = 2I- + S4O6(2-)
Find the moles of Iodine from second equation, then input it into first equation to find moles of oxidising agent ( divide by 3) and then can find concentration
