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Chemistry > Redox II > Flashcards

Redox II Flashcards

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1
Q

Oxidation in terms of electrons and oxidation number

A

Loss of electrons

Increase in oxidation number

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2
Q

Reduction in terms of electrons and oxidation number

A

Gain of electrons

Decrease in oxidation number

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3
Q

Always write redox equilibria int he form of

A

Reduction (electrons on LHS on equation)

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4
Q

What happens to electrons when a metal is placed in water?

A

Metal atoms shed electrons and the metal ions go into the water so that the electrons build up on the surface and result in a negative charge

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5
Q

How are positive ions attracted to a metal in water?

A

Buildup of electrons on the metal attracts positive ions in the solution to form a layer of positive ions

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6
Q

How do you some of the positive ions in solution form part of the metal?

A

Some of the positive ions in the layer regain that electrons from the buildup of electrons and become part of the metal surface again as a metal atom

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7
Q

How is a dynamic equilibrium establish when a metal is placed in water?

A

When the rate at which ions are leaving the surface = rate at which they are joining again (metal atoms reforming)

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8
Q

Explain what the diagrams for Magnesium in water and Copper in water as a snapshot of dynamic eqm look like?

A

Magnesium: lots of electrons on the surface and big layer of positive ions

Copper: less electrons on surface and less of a layer of positive ions (because Copper is less reaction so the Cu atoms shed electrons to form positive ions less easily)

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9
Q

Position of equilibrium in relation to how well something sheds slew from a and forms positive ions?

A

Readily sheds electrons and forms positive ions

Equilibrium lies further to the left

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10
Q

Standard Hydrogen electrode

Equilibrium
Pressure
Temperature
Concentration
Catalyst
A 
2H+(aq) + 2e-   H2(g)
Hydrogen pressure at 1 bar (100kPa)
298K
H2SO4 conc is 1mol dm^-3
Platinum catalyst
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11
Q

What is a cell?

A

The whole set up (two electrodes connected by a voltmeter and saltbridge)

Aka two half cells

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12
Q

What is a half cell?

A

Each of the two beakers and their contents

Ie. Mg electrode in beaker of Magnesium Sulphate

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13
Q

What is the purpose of a salt bridge?

A

Included to complete the electrical circuit without introducing any other pieces of metal into the system

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14
Q

What is a salt bridge?

A

Glass tube filled with an electrolyte like potassium nitrate solution

End stoppered by cotton wool to prevent mixing of contents in salt bridge with those in beaker

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15
Q

What will the system look like if a Magnesium electrode and platinum electrode for hydrogen are measured?

A

Magnesium electrode had much greater build up of electrons that the platinum electrode

Mg eqm lies further left

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16
Q

Explain why the voltmeter must be high resistance?

A

Avoids flow of current through the surface

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17
Q

What does E cell mean?

A

The electromotive (emf) of the cell is the maximum possible voltage in any situation.

18
Q

What is the cell diagram for when a magnesium electrode is coupled to a hydrogen electrode

And variants

A

Pt [H2(g)] | 2H+(aq) || Mg2+(aq) | Mg(s)

Pt | H2(g) | 2H+(aq) ||

Or

Pt | H2(g) | H+(aq) ||

19
Q

X

A

X

20
Q

Cell diagram conventions (3)

A

Draw an arrow from the E standard to the right hand electrode

[] Shows something flowing over a catalyst (eg. Hydrogen gas over platinum)

Substance losing electrons is written closest to its electrode

If more than one thing on either side of equilibrium (eg. With potassium dichromate(VI)) square brackets written around to keep tidy

21
Q

Define standard electrode potential

A

Emf measured when a metal electrode (or a metal ion electrode) is coupled to the standard hydrogen electrode under standard conditions

22
Q

Explain why equilibrium lying furthest to the left means the most negative E value?

A

Form ions more easily leaving electrons behind on the metal

making it more negative

23
Q

Connect E values, equilibrium direction and readily forming ions

A

Negative E value
Equilibrium further left
More readily element loses electrons to form ions
Stronger reducing agent

Positive E value
Equilibrium further right 
Less readily loses electrons/forms
ions/readily picks up electrons again
Stronger oxidising agent
24
Q

What is the electrochemical series?

A

Arranging various redox equilibria in order of standard electrode potentials

Most negative E value at top
Most positive E value at bottom

25
Q

Explain strong oxidising and reducing agents in relation to the electrochemical series?

A

Top: form ions more easily, good at giving away electrons so GOOD REDUCING AGENTS

Bottom: form ions less easily, good at picking up electrons so GOOD OXIDISING AGENTS

26
Q

Oxidising and reducing ability in relation to electrochemical series?

A

Oxidising ability of ions increases down the electrochemical series

Reducing ability of the metal increases up the electrochemical series

27
Q

Equilibrium lying more to the left means…

A

more likely to pick up electrons

28
Q

What happens if you have a system involving only gases?

Eg. Hydrogen and chlorine

A

Set up standard hydrogen electrode as normal

Set up chlorine gas bubbling over another platinum electrode immersed in 1 mol dm^-3 Cl- solution

29
Q

Explain how to measure redox potential for a system of Fe2+/Fe3+ ions

A

Platinum electrode inserted into beaker containing solution of iron (II) and iron (III) ions coupled with hydrogen electrode

30
Q

Potassium dichromate (VI) equilibrium

A

Cr2O72- (aq) + 14H+ (aq) + 6e- 2Cr3+ (aq) + 7H2O(l)

31
Q

Potassium dichromate half cell set up

A

Standard hydrogen electrode coupled with

Platinum electrode in beaker with solution containing Cr2O7 2-(aq), H+(aq) and Cr 3+(aq) ions at 1 mol dm^-3

32
Q

More negative E value means

A

Release electrons and forms ions more readily

33
Q

Zinc half cell and a copper half cell

A

You can combine two half cells not including the standard hydrogen electrode

The more negative E value = forms ions more readily = eqm further left compared to other half cell

Copper: less electrons and layer of positive ions
Zinc: more electrons and larger layer of positive ions

34
Q

Explain how to show the relationship between combinations of half cells and simple redox reactions carried out in test tubes

A

1) write redox equilibriums of both electrodes and E values
2) work out which E is the most -ve value so can work out which direction electrons flow
3) work out which way this means the eqm’s shift
4) write two one way redox equations
* see notes*

35
Q

Examples of test tube reactions that can be shown by redox equilibria electrodes without voltmeter (chemguide) (4)

A

1) Combining zinc half cell and copper half cell (e- flow from zinc to copper)
2) reaction between copper and silver nitrate (e- flow from copper to silver)
3) Reaction between magnesium and dilute sulphuric acid (e- flow from magnesium eqm to hydrogen eqm)
4) potassium dichromate (VI) oxidising Fe2+ —> Fe3+ (e- flow from Iron eqm to dichromate eqm)

36
Q

In whole cells where the voltmeter is removed

what is the rule of thumb for which way eqm shifts with E value?

A

Equilibrium with more negative E value shifts to left

Equilibrium with more positive E value shifts to the right

37
Q

How do you use an E value to predict the feasibility of a possible redox reaction?

A

1) write out the two redox equilibria and E values
2) find the most -ve E value to work out which direction the electrons flow. Then work out what effect this has on the shifting equilibrium.
3) check what things you start with and if the shift of equilibrium can happen versus if it can’t shift any more in one direction to work out feasibility/non feasibility

38
Q

Which redox reaction is predicting the feasibility difficult for depending on what reacts?

A

Copper and dilute nitric acid

Copper could react with either H+ (not feasible because eqm already shifted as far as possible)
or
NO3 - (feasible) and produces Cu2+ and NO2

39
Q

Why might an E value predicting feasibility appear incorrect?

X

A

Not feasible because large activation energy barrier

40
Q

Name two examples where E values predicting feasibility appear wrong and say why.

A

1) acidified potassium dichromate (VI) oxidising water.
E value says it is feasible but nothing happens in the test tube because a large activation energy barrier must be overcome

2) acidified potassium dichromate (VI) oxidising chloride ions to chlorine
E value says it is not feasible however it is feasible if you change standard conditions using HCL to give H+ and Cl- ions at 10 mol dm^-3 concentration

41
Q

Explain how to use redox equilibria to select an oxidising agent

A

Write out redox equilibria, in reduction form, and E value for thing you want to oxidise
Eg. Fe3+(aq) + e- Fe2+(aq). E = +0.77v

To oxidise equilibrium must shift left (E value for redox equilibrium you want to shift must be more negative)

So find another equilibrium you can couple this with that has an E value that is less negative than the eqm one you already have

42
Q

Explain how to use redox equilibria to select a reducing agent

A

Write out redox equilibria, in reduction form, and E value for thing you want to reduce
Eg. Cr3+(aq) + e- Cr2+(aq). E = -0.41

To reduce, equilibrium must shift right (E value for redox equilibrium you want to shift must be less negative)

So find another equilibrium you can couple this with that has an E value that is more negative than the eqm one you already have

Chemistry (37 decks)

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