Redox II

Question 1

Oxidation in terms of electrons and oxidation number

Answer 1

Loss of electrons

Increase in oxidation number

Question 2

Reduction in terms of electrons and oxidation number

Answer 2

Gain of electrons

Decrease in oxidation number

Question 3

Always write redox equilibria int he form of

Answer 3

Reduction (electrons on LHS on equation)

Question 4

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

Answer 4

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

Question 5

How are positive ions attracted to a metal in water?

Answer 5

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

Question 6

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

Answer 6

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

Question 7

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

Answer 7

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

Question 8

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

Answer 8

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)

Question 9

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

Answer 9

Readily sheds electrons and forms positive ions

Equilibrium lies further to the left

Question 10

Standard Hydrogen electrode

Equilibrium
Pressure
Temperature
Concentration
Catalyst

Answer 10

2H+(aq) + 2e-   H2(g)
Hydrogen pressure at 1 bar (100kPa)
298K
H2SO4 conc is 1mol dm^-3
Platinum catalyst

Question 11

What is a cell?

Answer 11

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

Aka two half cells

Question 12

What is a half cell?

Answer 12

Each of the two beakers and their contents

Ie. Mg electrode in beaker of Magnesium Sulphate

Question 13

What is the purpose of a salt bridge?

Answer 13

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

Question 14

What is a salt bridge?

Answer 14

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

Question 15

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

Answer 15

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

Mg eqm lies further left

Question 16

Explain why the voltmeter must be high resistance?

Answer 16

Avoids flow of current through the surface

Question 17

What does E cell mean?

Answer 17

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

Question 18

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

And variants

Answer 18

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

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

Or

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

Question 19

X

Answer 19

X

Question 20

Cell diagram conventions (3)

Answer 20

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

Question 21

Define standard electrode potential

Answer 21

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

Question 22

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

Answer 22

Form ions more easily leaving electrons behind on the metal

making it more negative

Question 23

Connect E values, equilibrium direction and readily forming ions

Answer 23

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

Question 24

What is the electrochemical series?

Answer 24

Arranging various redox equilibria in order of standard electrode potentials

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

Question 25

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

Answer 25

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

Question 26

Oxidising and reducing ability in relation to electrochemical series?

Answer 26

Oxidising ability of ions increases down the electrochemical series Reducing ability of the metal increases up the electrochemical series

Question 27

Equilibrium lying more to the left means...

Answer 27

more likely to pick up electrons

Question 28

What happens if you have a system involving only gases? | Eg. Hydrogen and chlorine

Answer 28

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

Question 29

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

Answer 29

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

Question 30

Potassium dichromate (VI) equilibrium

Answer 30

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

Question 31

Potassium dichromate half cell set up

Answer 31

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

Question 32

More negative E value means

Answer 32

Release electrons and forms ions more readily

Question 33

Zinc half cell and a copper half cell

Answer 33

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

Question 34

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

Answer 34

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*

Question 35

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

Answer 35

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)

Question 36

In whole cells where the voltmeter is removed what is the rule of thumb for which way eqm shifts with E value?

Answer 36

Equilibrium with more negative E value shifts to left Equilibrium with more positive E value shifts to the right

Question 37

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

Answer 37

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

Question 38

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

Answer 38

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

Question 39

Why might an E value predicting feasibility appear incorrect? X

Answer 39

Not feasible because large activation energy barrier

Question 40

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

Answer 40

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

Question 41

Explain how to use redox equilibria to select an oxidising agent

Answer 41

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

Question 42

Explain how to use redox equilibria to select a reducing agent

Answer 42

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