Standard electrode (redox) potentials

Question 1

in terms of electrons, oxidation is

Answer 1

the loss of electrons

Question 2

in terms of electrons, reduction is

Answer 2

the gain of electrons

Question 3

in terms of change in oxidation number, oxidation is

Answer 3

when the oxidation number increases

Question 4

in terms of change in oxidation number, reduction is

Answer 4

when the oxidation number decreases

Question 5

the reactions involved in the measurements of a standard electrode potential are……………..reactions in a state of……………….

Answer 5

redox reactions in equilibria

Question 6

when a metal such as Magnesium or Copper, represented by ‘M’, is placed in water, there is a very small tendency for the metal atoms to…..

Answer 6

lose electrons and go into solution as positive metal ions

Question 7

write the equations that represents the loss of electrons from either Magnesium or Copper atoms (‘M’ and ‘x’ for the charge) to form aqueous metal ions and electrons: (write the non-reversible equation and the reversible one once equilibrium has been reached)

Answer 7

M(s) → xe- + Mx+(aq)

Mx+(aq) + xe- → M(s)

xe- + Mx+(aq) ⇌ M(s)

Question 8

explain the following equations in relation to the establishment of an equilibrium:
M(s) → xe- + Mx+(aq)
Mx+(aq) + xe- → M(s)
xe- + Mx+(aq) ⇌ M(s)

Answer 8

M(s) → xe- + Mx+(aq)
-The metal atoms lose electrons and go into solution as positive metal ions, with the electrons remaining on the metal’s surface.
-the electrons build up on the metal’s surface and the resulting negative charge attracts the positive metal ions in solution, creating a layer of positive ions surrounding the metal
Mx+(aq) + xe- → M(s)
-some of the metal onis will regain their electrons from the surface of the metal and return to form part of the metal
xe- + Mx+(aq) ⇌ M(s)
-eventually, a dynamic equilibrium is reached as the rate at which metal atoms leave the metal surface is the same as the rate at which metal ions join the metal from solution

Question 9

• the equations showing the tendencies of magnesium (Mg) and copper (Cu) to release electrons to form positive ions are:
• also show the position of equilibrium for each!

Answer 9

Mg2+(aq) + 2e- ⇌ Mg(s)
Mg’s position of equilibrium lies further to the left than for copper, because of magnesium’s greater tendency to release electrons than copper. this means magnesium will have a greater negative charge on the metal surface and more positive ions in solution
Cu2+(aq) + 2e- ⇌ Cu(s)

Question 10

the absolute potential difference is

Answer 10

the potential difference between the metal and the solution

Question 11

it is not possible to measure the absolute potential difference because:

Answer 11

-though it is easy to connect the metal electrode to a voltmeter, the second electrode would also have to be put into the solution, but would create its own potential difference instead of measuring the solution’s potential difference with the metal

Question 12

the standard hydrogen electrode consists of

Answer 12

hydrogen gas at a pressure of 100 kPa (1 bar) bubbling over a piece of platinum foil dipped into a solution of hydrochloric acid (or sulfuric acid) with a hydrogen ion concentration of 1 ol dm-3 at a temperature of 298 K

Question 13

in the standard hydrogen electrode, the platinum foil is covered in ………………………. which allows an equilibrium between the hydrogen ions in solution and the hydrogen gas to be established quickly because……………….

Answer 13

covered in porous platinum

porous platinum has a large surface area so allows the equilibrium to be reached more quickly

Question 14

the standard hydrogen electrode equilibrium equation is

Answer 14

H+(aq) + e- ⇌ 1/2H2(g)

Question 15

when writing equilibria for standard electrode potentials, we ALWAYS write the electrons on the ………….. hand side

Answer 15

LEFT hand side

Question 16

the standard conditions that apply to all equilibria so that a fair comparison can be made between them are:

Answer 16

gas pressure, 100 kPa
temperature, 298K
concentration of ions in solution, 1 mol dm-3

Question 17

describe how you would measure the standard electrode potential of a metal ion | metal system such as Mg2+ | Mg

Answer 17

• set up a standard hydrogen electrode by bubbling H2 gas over a piece of platinum foil at 100 kPa pressure and 298 K
• use a solution of HCl acid at a 1 mol dm-3 H+ ion concentration and use porous platinum to cover the platinum foil
• clean a piece of magnesium metal with sand paper and attach attach to a connecting wire. also attach a connecting wire to the platinum foil and link both up to a high resistance voltmeter
• dip the Mg electrode into a 1 mol dm-3 concentration magnesium sulfate solution
• link up both beakers with a salt bridge made of filter paper saturated with potassium nitrate (KNO3)

Question 18

why is a high resistance voltmeter used?

Answer 18

ideally this would have infinite resistance so there would be no flow of electrons (no current) around the external circuit so that the reading would represent the difference in potential between the two half cells when both reactions are at equilibrium.

Question 19

why is an ionic salt used in the salt bridge, and why is it usually KNO3 ?

Answer 19

the ionic salt allows the movement of ions between the half-cells and its usually KNO3 because the salt bridge ionic salt cannot interact with any of the half-cell ions, and usually KNO3 is suitable

Question 20

E⦵cell =

Answer 20

E⦵oxidised - E⦵reduced

Question 21

the standard electrode potential, E⦵ , of the standard hydrogen electrode is

Answer 21

E⦵ = 0.00 V

Question 22

the sign ( + of -) of the electrode potential, E⦵, indicates

Answer 22

the polarity of the electrode relative to the hydrogen electrode. this sign is fixed and does not change even if the equation of the half cell is reversed

Question 23

the standard electrode potential are measured when

Answer 23

no electrons are flowing and hence when both half-cell reactions are in equilibrium

Question 24

the E⦵ value for Mg2+(aq) | Mg(s) is -2.37V while the value for Cu2+(aq) | Cu(s) is +0.34V. what do the signs on the E⦵ values tell us about the relative reducing and power of magnesium and copper

Answer 24

the negative value for Mg2+(aq) | Mg(s) tells us the equilibrium position if further tot he left than the equilibrium position. the positive sign for Cu2+(aq) | Cu(s) indicates the position of equilibrium of this reaction of further to the right than the equilibrium positions of the reaction in the hydrogen electrode.
therefore, magnesium releases electrons more readily than hydrogen and that copper releases electrons less readily than hydrogen.
this means magnesium is a better reducing agent than both hydrogen and copper, while hydrogen is better reducing agent than copper

Question 25

a reducing agent is

Answer 25

a species that reduces another species by adding on or more electrons to it, so is therefore itself oxidised

Question 26

a negative E⦵ means that

Answer 26

the position of equilibrium of the half cell reaction lies FURTHER TO THE LEFT than the equilibrium of the standard hydrogen electrode

Question 27

a positive E⦵ means

Answer 27

the position of equilibrium of the half cell reaction lies FURTHER TO THE RIGHT than the equilibrium of the standard hydrogen electrode

Question 28

the electromotive force (emf), E⦵cell, is

Answer 28

the standard electrode potential of a half-cell (measured under standard conditions of 298 K, 100 kPa pressure and concentration of 1 mol dm-3) connected to a standard hydrogen electrode

Question 29

the emf value of a cell in which the hydrogen electrode is the positive electrode will have a ………value

Answer 29

negative value of E⦵cell

Question 30

the emf value of a cell in which the hydrogen electrode is the negative electrode will have a ………value

Answer 30

positive value of E⦵cell

Question 31

here are the E⦵ values for two reactions:
Li+(aq) + e- ⇌ Li(s) -3.03V
1/2F2 + e- ⇌ F-(aq) +2.87V
use these values to determine which species is most reducing and which is most oxidising

Answer 31

lithium is the most reducing agent because its E⦵ value is most negative, meaning it can lose electrons more readily than F.
fluorine is the most oxidising species because it can gain electrons most readily due to its least negative E⦵ value