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A loss of electrons is called
A gain in electrons is called
When reduction and oxidation happen simultaneously it is called a
An oxidising agent ____ electrons and gets ____
A reducing agent ____ electrons and gets ____
There are 8 rules for oxidising:
- All atoms are treated as ___ for this, even if they are covalently bonded
- Uncombined elements have an oxidation number of
- Elements just bonded to indentical atoms have a oxidatiom number of
- The oxidation number of simple monotomic ion, is the same as its
- In compounds/compound ions, the overall oxidation number is just the
- Thesum of oxidation numbers for neutral componds is
- Combined oxygen is nearly always __ exept in peroxides where it's ___. And in OF2 is __ and O2F2 where it is __
- Combined H is __ expect in metal hydrides where its __
- ionic charge
Roman numerals give oxidation numbers.
What is the oxidation state of Cu in Copper (||) Sulphate
Oxidation state for an atom will ____ by 1 for each election lost and vise versa.
An element being oxidized and reduced at the same time is called
Zn displaces Ag ions from silver nitrate solution to form zink nitrate and a deposite of silver metal.
Give the ionic half equations and the full ionic equation for this reaction.
- Zn → Zn2+ + 2e-
- Ag+ + e- → Ag
- Zn + 2Ag+ → 2Ag + Zn2+
Cl reacts with NaOH solition to give a mixture of NACl and NaClO and H20. The balanced reaction is:
2NaOH + Cl2 → NaCl + NaClO + H20
What are the oxidation numbers for Cl in NaCl and NaClO formed during the reaction?
- NaCl has an oxidation number of -1
- NaClO has an oxidation number of +1
Two ionic equations are:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
2I- → I2 + 2e-
a) Write a balanced equation to show the reaction taking place.
b) Use oxidation numbers to explain the redox processes which have occured
c) Suggest why a fairly reactive metal such as Zn will not react with I-(aq) in a similar manner to manganate (VII) ions.
- a) 2MnO4- + 16H+ 10I- → 2Mn + 8H2O + 5I2
- b) Mn in MnO4- has an oxidation number of +7 which is reduced to +2 where as I- is oxidsed from -1 to 0
- c) Recative metals have a tendancy to be able to lose electrons easily, meaning that they are good reducing agents. As I- is already reduced it wouldn't react.
Electochemical cells can be made from what?
Two different metals dipped in salt solutions of their own ions and connected by a wire (the external circuit)
Theres always two recations within an electrochemical cell. So its a what reaction?
Draw a detailed sketch of electochemical cell of Zn and Cu and the half equations which occur.
This shows that electrons flow through the wire in which direction?
From the most reactive metal to the least reactive
You can also have half-cells involving soltuions of 2 aqueous ions of the same element. Draw this for Fe2+/Fe3+
Rev guide page 74
Electrode potentials are measured against a
Standard Hydrogen Electrode
What is the standard electrode potential of a half cell?
The standard electrode potential of a half-cell is the voltage measured under standard conditions when the half-cell is connected to a standard hydrogen electrode.
A standard hydrogen electrode contains what?
The equilibrium which is set up
And it value
- A peice of Pt foil submerged in a 1 mol dm-3 solution of H+ ions.
- H2 gas is bubbled through at a pressure of 101 kPa.
- The Pt electrode is platinized - which means the surface areaof the foil is covered in ve
- ry finely powdered Pt, increasing its surface area.
- The platinized surface absorbs hydrogen gas and an equilibrium is set up:
It is defined as 0 and is known as the reference cell
- It can be shown on the diagram:
The SHE is always on what side?
Eo right handside - Eo left handside
The more negative electrode is on the
Left hand side
To write the overall cell equation, you would combine the half-equations.
To do this, you would...
- Write the oxidised substance first
- So the first reaction goes backwards
- Bottom reaction goes forwards
- Combining these two give the overall equation
What is the Eocell of the following reaction:
Fe2+ + 2e- ⇄ Fe -0.44V
Pb2+ + 2e- ⇄ Pb -0.13V
(-0.13) - (-0.44) = +0.31V
Zn2+ + 2e- ⇄ Zn - 0.76 V
Ag+ + e- ⇄ Ag + 0.80 V
a) Use the standard electrode potentials given to calculate the standard cell potential for a zinc-silver cell.
b) Write an equation for the overall cell reaction
c) Which half-cell released the electrons into the circuit. Explain your answer.
- a) 0.80 - (-0.76) = 1.56 V
- b) Zn + 2Ag+ ⇄ Zn2+ + 2Ag
- c) Zinc because it's being oxidised. Meaning it is losing its electrons into the circuit where as Ag is being reduced.
The more reactive a metal is, the more it want to...
Lose electrons to form a positive ion
More reactive metals have..
More negative standard electrode potentials
The more reactive a non-metal is, the more it wants to gain electrons to form a negative ion. More reactive non-metals have...
More positive standard electrode potentials
If the Eo is x then the reaction would be:
a) < -0.6
b) Approx. -0.1
d) Approx. +0.1
e) > +0.6
- a) Reaction does not take place
- b) Reaction to the left
- c) Equal amounts of reactants/products
- d) Reaction to the right
- e) Reaction to completion
Predictions of the value of the Eo cell may be wrong. This is because:
The conditions are not standard, such as changing the concentraion or temperature of the cell. It would affect them as they are in equilibrium
The reaction kinetics are not favourable. The Ea might be high which may stop it from happening. Or that the reaction is too slow that the reaction might not appear to happen.
V is a transition element with several oxidation states, which are differently coloured. Vandium(V)oxide is yellow and vandium(VI)oxide is blue
Zn2+ + 2e- ⇄ Zn -0.76V
VO2+ + 2H+ + e- ⇄ VO2+ + H2O 1.00V
a) Combine the two half equations and predict if Zn will convert V from its yellow to blue state.
b) Calculate the cell potential for this reaction.
- a) 2VO2+ + 4H+ + Zn ⇄ 2VO2+ + Zn2+ + 2H2O
- As VO2+ goes to VO2+ it does change from yellow to blue.
- b) 1.76 V
Titrations using transition element ions are
Titrations using transition element ions let you find out how much oxidising agent is needed to exactly react with a quantity of reducing agent. If you know the concentration of either the oxidising agent or the reducing agent, you can use the titration results to work out the concentration of the other. How would you do this?
- 1. Measure out a quantity of a reducing agent using a pipette and put it in a conical flask.
- 2. Using a measuring cylinder, you add about 20cm3 of the dilute sulphuric acid to the flask. This is in excess, so you don't need to be too exact.
- 3. Now you add the oxidising agent to the reducing agent using a burette swirling the conical flask as you do so
- 4. Stop when the mixture in the flask becomes tainted with the colour of the oxidising agent and record the volume of the oxidising agent added. This is the rough titration.
- 5, Now you do some more accurate titrations. You need to get readings that are within 0.1cm3 of each other
When the coloured oxidising agent is added to the reducing agent, they start reacting and the reducing agent starts to
Lose its colour
The reaction continues until all of the reducing agent has reacted. The very next drop into the flask will give the mixture the
Colour of the oxidising agent
What are two main oxiding agents which are used?
- 1. Manganate (VII) ions (MnO4-) in aqueous potassium manganate (VII) (KMnO4) These are purple.
- 2. Dichromate (VI) ions (Cr2O72-) in aqueous potassium dichromate (VI) (K2Cr2O7) These are orange.
The acid is added because...
It is to make sure that there are plenty of H+ ions to allow the oxidising agent to be reduced.
A 3.2g Fe tablet was dissolved in dilute sulphuric acid and made up to 250 cm3 with deionized water. 25 cm3 of this soln was found to react with 7.5 cm3 of 0.018 mol dm-3 potassium manganate (VII) soln
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + Fe3+
a) Calculate the number of moles of Fe in 25 cm3 of the solution
b) Calculate the number of moles of Fe in the tablet
c) What %, by mass, of the tablet is Fe?
- a) (0.018 × 7.5) ÷ 1000 = 1.35 x 10-4
- 1.35 x 10-4 x 5 = 6.75 x 10-4 moles
b) 6.75 x 10-3
- c) 56 x 6.75 x 10-3 = 0.378
- (0.378 ÷ 3.2) x 100 = 11.8%
A 10cm3 sample of 0.5 mol dm-3 SnCl2 solution was titrated with acidified