Electrochemistry

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 metalMx+(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 metalxe- + 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 solutionCu2+(aq) + 2e- ⇌ Cu(s)

Question 10

<p>Define standard electrode potential </p>

Answer 10

<p>the potential difference between the metal and the solution</p>

Question 11

the standard hydrogen electrode consists of

Answer 11

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 12

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 12

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

Question 13

the standard hydrogen electrode equilibrium equation is

Answer 13

H+(aq) + e- ⇌ 1/2H2(g)

Question 14

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

Answer 14

LEFT hand side

Question 15

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

Answer 15

gas pressure, 100 kPatemperature, 298Kconcentration of ions in solution, 1 mol dm-3

Question 16

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

Answer 16

-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 17

why is a high resistance voltmeter used?

Answer 17

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 18

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

Answer 18

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 19

<p>E⦵cell =</p>

Answer 19

<p>E⦵red - E⦵ox</p>

Question 20

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

Answer 20

E⦵ = 0.00 V

Question 21

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 21

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 22

a reducing agent is

Answer 22

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

Question 23

a negative E⦵ means that

Answer 23

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

Question 24

a positive E⦵ means

Answer 24

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

Question 25

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

Answer 25

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 26

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

Answer 26

negative value of E⦵cell

Question 27

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

Answer 27

positive value of E⦵cell

Question 28

here are the E⦵ values for two reactions:Li+(aq) + e- ⇌ Li(s) -3.03V1/2F2 + e- ⇌ F-(aq) +2.87Vuse these values to determine which species is most reducing and which is most oxidising

Answer 28

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

Question 29

which of the following will be the negative electrode of the cell:Zn2+(aq) + 2e- ⇌ Zn(s)E⦵ = -0.76VCu2+(aq) + 2e- ⇌ Cu(s)E⦵ = +0.34V

Answer 29

Zn2+(aq) + 2e- ⇌ Zn(s)E⦵ = -0.76VCu2+(aq) + 2e- ⇌ Cu(s)E⦵ = +0.34Vthe E⦵ value for the Zn2+ | Zn half-cell is more negative, so the zinc electrode will be the negative electrode of the cellso the electrons will flow from the zinc electrode to the copper electrode

Question 30

the cell-diagram for the cell formed by combining the half cells below are:Zn2+(aq) + 2e- ⇌ Zn(s)E⦵ = -0.76VCu2+(aq) + 2e- ⇌ Cu(s)E⦵ = +0.34V

Answer 30

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Question 31

use the following two equilibria and their E⦵ values to how relatively their equilibria are far left to the standard hydrogen electrode:Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 VCu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 31

The position of equilibrium of the Zn2+(aq) | Zn(s) reaction lies further to the left than that of the Cu2+(aq) | Cu(s) reaction as the Zn2+(aq) | Zn(s) E⦵ value if more negative than the Cu2+(aq) | Cu(s) E⦵ value

Question 32

if the following equilibria are linked by combining the two half-cells to make an electrochemical cell, the electrons will flow from the …………..electrode to the …………….electrodeZn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 VCu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 32

e- flow from the zinc electrode to the copper electrode because zinc is the negative electrode, having the more negative E⦵ value for the half-cellelectrons flow from the half-cell with the more negative E⦵ value to the half-cell with the less negative E⦵ value

Question 33

explain whether the following reaction thermodynamically feasible:Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)use the following info to help you: Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 VCu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 33

-The E⦵ value for the Zn equilibrium is more negative than the Cu equilibrium (-0.76V to +0.34V)-so the position of the Zn equilibrium will shift to the left, releasing electrons while the position of the Cu will shift to the right, accepting electrons-this means the reaction between Zn(s) and Cu2+(aq) is thermodynamically feasible

Question 34

use the data below to deduce whether zinc will react with dilute sulfuric acid:Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 V2H+(aq) + 2e- ⇌ H2(g) E⦵ = 0.00 V

Answer 34

• with the half equations arranged in order of reducing power (most negative on top, most positive on bottom), you can see Zn(s) reacts with H+• therefore the equation for the reaction is:Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)• calculating the E⦵cell value for the reaction gives:E⦵cell = E⦵red - E⦵oxi (E⦵RHS - E⦵LHS)E⦵cell = 0.00 - -0.76 = +0.76V• Zinc releases e- to the hydrogen ions• The Reaction is thermodynamically feasible

Question 35

use the data below to deduce whether copper will react with dilute sulfuric acid:2H+(aq) + 2e- ⇌ H2(g) E⦵ = 0.00 V Cu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 35

• With the half equation arranged in order of reducing power, we can see that Cu(s) will NOT react with H+ ions as the copper equilibrium is more positive, meaning:Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) is not feasible because:E⦵cell = 0.00 - 0.34 = -0.34 V• E⦵cell is negative, so the reaction is not thermodynamically feasibleinstead, the reaction:Cu2+ + H2(g) → Cu(s) + 2H+(aq) is feasible, but has a very large activation energy, so the reactants are said to be kinetically stable