Chem ch26

Worked solutions to student book questions Chapter 26 Electricity from chemical reactions

Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. Draw a labelled diagram of a cell formed from Cl2/Cl and Sn2+/Sn half cells. Use the electrochemical series to indicate: a half cell reactions b the anode and the cathode c direction of electron flow d electrode polarities (which electrode is positive and which is negative) e the overall reaction

A1.

Q2. Repeat Question 1 for the cells formed from the following half cells: a Fe3+/Fe2+ and H+/H2 b Cl2/Cl and Pb2+/Pb

A2.

a



b

Q3. Show that under standard conditions, according the electrochemical series, the Daniell cell (Figure 26.2) should have a cell potential difference of 1.10 V.

A3.

cell potential difference = higher half cell E lower half cell E = E (Cu2+(aq)/Cu(s)) E (Zn2+(aq)/Zn(s)) = 0.34 (0.76) = 1.10 V

E1. The equation for the Daniell cell shown in Figure 26.2 is:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) The Nernst equation for this reaction is:

Ecell = Cu/Cu 2+E Zn/Zn2+E 2059.0 log

)]aq(Cu)][s(Zn[)]s(Cu)][aq(Zn[

2

2

+

+

Calculate the cell potential if: a [Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 1.0 M b [Zn2+(aq)] = 103 M and [Cu2+(aq)] = 1.0 M c [Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 103 M



AE1.

Ecell = E Cu2+/Cu E Zn2+/Zn 2059.0 log

2

2

[Cu(s)][Zn (aq)][Cu (aq)][Zn(s)]

+

+

From a table of standard electrode potentials: E Cu2+/Cu = 0.34V, and E Zn2+/Zn = 0.76 V. The concentration of solids are assigned a value of 1. The Nernst equation for the Daniell cell simplifies to

Ecell = 0.34 (0.76) 2059.0 log

2

2

[Zn (aq)][Cu (aq)]

+

+ or

Ecell = 1.10 2059.0 log

2

2

[Zn (aq)][Cu (aq)]

+

+

a Substituting the values for the concentration of the ions into the equation gives:

Ecell = 1.10 2059.0 log

0.10.1

Since log 1 = 0, Ecell = 1.10 V

b substitute [Zn2+(aq)] = 103 M and [Cu2+(aq)] = 1.0 M into the equation gives

Ecell = 1.10 2059.0 log

0.1 10 3

= 1.10 2059.0 log103

= 1.10 2059.0 (3)

= 1.10 + 0.0885 = 1.1885 = 1.2 V

c Substitute [Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 103 M into the equation

Ecell = 1.10 2059.0 log

2

2

[Zn (aq)][Cu (aq)]

+

+

= 1.10 2059.0 log 310

1

= 1.10 2059.0 3

= 1.10 0.0885 = 1.0115 = 1.01 V



Q4. The following equations appear in the order shown in the electrochemical series:

Cl2(g) + 2e 2Cl(aq) Ag+(aq) + e Ag(s) Zn2+(aq) + 2e Zn(s)

i Predict whether a redox reaction would occur in the following mixtures. Assume 1 M and 1 atm.

For mixtures where you predict a reaction will occur: ii write separate half equations for the oxidation and reduction reactions iii write a balanced equation. a Cl2(g) and Zn(s) b Ag+(aq) and Ag(s) c Ag+(aq) and Zn(s) d Zn2+(aq) and Cl(aq)

A4. a i Reaction will occur.

ii Zn(s) Zn2+(aq) + 2e (oxidation) Cl2(g) + 2e 2Cl(aq) (reduction)

iii Zn(s) + Cl2(g) Zn2+(aq) + 2Cl(aq) b i no reaction c i Reaction will occur.

ii Zn(s) Zn2+(aq) + 2e (oxidation) 2Ag+(aq) + e 2Ag(s) (reduction)

iii 2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+(aq) d i no reaction



Chapter review

Q5. Iron nails are placed into 1 M solutions of CuSO4, MgCl2, Pb(NO3)2 and ZnCl2. In which solutions would you expect a coating of a metal other than iron to appear on the nail? Explain your answer.

A5. Coatings of metals other than iron would be expected on the nails placed in 1 M solutions of CuSO4 and Pb(NO3)2. These solutions contain oxidants that are strong enough to react with iron metal. The iron in turn reduces the metal ions in the solution, forming a coating on the nail.

Q6. Use the electrochemical series to predict whether a reaction will occur in each of the following situations. Write an equation for each reaction that you predict will occur. a copper(II) sulfate solution is stored in an aluminium container b sodium chloride solution is stored in a copper container c silver nitrate solution is stored in a zinc container d an iron nail is placed in 1 M hydrochloric acid solution e a plumber uses hydrochloric acid to clean copper pipes

A6.

a 3Cu2+(aq) + 2Al(s) 3Cu(s) + 2Al3+(aq) b no reaction c 2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+(aq) d 2H+(aq) + Fe(s) H2(g) + Fe2+(aq) e no reaction

Q7. Explain the difference between the terms: a oxidant and reductant b anode and cathode c conjugate redox pair and conjugate acidbase pair d external circuit and internal circuit

A7. a Oxidant: a substance that causes another substance to be oxidised and is reduced

in the process. Reductant. a substance that causes another substance to be reduced and is oxidised in the process.

b anode: electrode at which oxidation occurs cathode: electrode at which reduction occurs

c conjugate redox pair an oxidant and its corresponding reduced form, e.g. Cu2+/Cu. The reduced form has gained electrons. Conjugate acid base pair: A conjugate acid contains one more hydrogen ion or proton than the base.



d external circuit: section of a circuit where the electrons flow e.g. through wires internal circuit: part of cell where the current is due to the movement of ions, e.g. in the salt bridge.

Q8.

The following equations form part of the electrochemical series. They are ranked in the order shown.

Ag+(aq) + e Ag(s) Pb2+(aq) + 2e Pb(s) Fe2+(aq) + 2e Fe(s) Zn2+(aq) + 2e Zn(s) Mg2+(aq) + 2e Mg(s)

a Which species is the strongest oxidant and which is the weakest oxidant? b Which species is the strongest reductant and which is the weakest reductant? c Lead rods are placed in solutions of silver nitrate, iron(lI) sulfate and magnesium

chloride. In which solutions would you expect to see a coating of another metal form on the lead rod? Explain.

d Which of the metals silver, zinc or magnesium might be coated with lead when immersed in a solution of lead(II) nitrate?

A8. a Ag+(aq); Mg2+(aq) b Mg(s); Ag(s) c A coating of silver will form on the lead when it is placed in silver nitrate

solution because Ag+ ions are stronger oxidants than Pb2+ ions. d zinc and magnesium

Q9. Use the electrochemical series to determine whether: a elemental iodine is an oxidant or reductant b calcium metal is a strong or weak reductant c the permanganate ion is a strong or weak oxidant d nickel is a better reductant than silver e Cu2+(aq) is a better oxidant than Ag+(aq) f Fe2+(aq) can act as an oxidant and a reductant

A9. a oxidant b strong reductant c strong oxidant d true e false f true



Q10. a Use the electrochemical series to predict what might be expected to occur if

hydrogen gas were bubbled through a solution containing Fe3+ ions. b Write an equation for the predicted reaction. c When the reactants were mixed in an experiment, no reaction was observed.

Suggest possible reasons for this.

A10. a predict Fe2+(aq) and H+(aq) are formed b 2Fe3+(aq) + H2(g) 2Fe2+(aq) + 2H+(aq) c If significant reaction had occurred, the yellow solution containing Fe3+ ions

would have become pale green as Fe2+ ions formed. Since no reaction was observed, the rate of the reaction may have been slow. Alternatively, it must be remembered that the electrochemical series is only valid for certain conditions. It is possible that under the conditions in which the experiment was performed, little reaction would occur.

Q11. A student was told that the redox pair Mn2+(aq)/Mn(s) is lower in the electrochemical series than the Fe2+(aq)/Fe(s) pair and higher than Al3+(aq)/Al(s). Explain how the student could experimentally determine the exact location of the manganese pair in the series on page 418.

A11. By referring to the electrochemical series, the student will see that the position of the Mn2+/Mn pair in the series must be determined in relation to the Zn2+/Zn pair and the H2O/H2, OH pair. The student could construct an electrochemical cell from an Mn2+/Mn half cell and a Zn2+/Zn half cell. If the manganese electrode was found to be the positive electrode, then Mn2+ would be a stronger oxidant than Zn2+ and the Mn2+/Mn pair should be placed between the Zn2+/Zn pair and the Fe2+/Fe pair in the series. If the manganese electrode were negative, then Mn2+ would be a weaker oxidant than Zn2+ and the Mn2+/Mn pair should be placed below the Zn2+/Zn pair. In this case, the student could determine whether to place the manganese pair above or below the H2O/H2, OH pair by constructing a cell from an Mn2+/Mn half cell and an H2O/H2, OH half cell and finding the polarity of the manganese electrode.

Q12. Account for the following observations. a Bromine reacts with iodide ions in solution but does not react with chloride ions. b Hydrogen peroxide (H2O2) solution can spontaneously decompose to form water

and oxygen. c Tin metal is added to solutions of tin(II) chloride to prevent oxidation of the

tin(II) ions by oxygen in air. d Blocks of zinc are attached to the iron hulls of ships to reduce corrosion.



A12. a Bromine is a stronger oxidant than iodine, so it reacts with iodide ions. Since

bromine is a weaker oxidant than chlorine, bromine does not react with chloride ions.

b Hydrogen peroxide acts as both a strong oxidant and as a weak reductant. (It appears in the electrochemical series on the left side of one half reaction and the right side of another.) Hydrogen peroxide therefore reacts with itself. The reaction is very slow unless a catalyst such as manganese dioxide is added.

c Sn2+ ions can be oxidised to Sn4+ ions by a suitable oxidant. Tin metal can reduce Sn4+ ions so that they re-form Sn2+ ions.

d Zinc is a stronger reductant than iron and, if the two metals are in contact, the zinc is oxidised preferentially. For this reason, the presence of a zinc block on the iron hull of a ship protects the hull from corrosion.

Q13. A student working in the laboratory spilled an iodine solution over the bench, causing a dark brown stain to form. Suggest how the student could remove the iodine stain.

A13. The brown stain contains iodine (I2). A reductant stronger than I ions would react with iodine. By referring to the electrochemical series, you can see that a solution containing Sn2+ ions might react and cause the stain to be removed.

Q14. What are the limitations that need to be considered when using the electrochemical series to predict whether or not a certain reaction will occur?

A14. The electrochemical series is based on reactions occurring under standard conditions of 298 K, 1 atmosphere pressure and 1 M concentration and can only be used to predict the possibility of reaction occurring under these conditions. The series give no information about the rate of reactions so even if a reaction is predicted it may be so slow that no reaction is apparent.

Chem ch26

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