Chapter 26 Electricity from Chemical Reactions

Worked solutions to textbook questions1

Chapter 26 Electricity from chemical reactions

Q1.

Draw a labelled diagram of a cell formed from Cl2/Cl– and Sn2+/Sn half cells. Use the electrochemical series to indicate:

ahalf cell reactions

bthe anode and the cathode

cdirection of electron flow

delectrode polarities (which electrode is positive and which is negative)

ethe overall reaction

A1.

Q2.

Repeat Question 1 for the cells formed from the following half cells:

aFe3+/Fe2+ and H+/H2

bCl2/Cl– and Pb2+/Pb

A2.

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 E0– lower half cell E0

= E0 (Cu2+(aq)/Cu(s)) – E0 (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 = – – log

Calculate the cell potential if:

a[Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 1.0 M

b[Zn2+(aq)] = 10–3 M and [Cu2+(aq)] = 1.0 M

c[Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 10–3 M

AE1.

Ecell = E0 Cu2+/Cu – E0 Zn2+/Zn – log

From a table of standard electrode potentials:

E0 Cu2+/Cu = 0.34V, and E0 Zn2+/Zn = –0.76V.

The concentration of solids are assigned a value of 1.

The Nernst equation for the Daniell cell simplifiesto

Ecell = 0.34 – (–0.76) – log or

Ecell = 1.10 –log

aSubstituting the values for the concentration of the ions into the equation gives:

Ecell = 1.10 –log

Since log 1 =0, Ecell = 1.10V

bsubstitute [Zn2+(aq)] = 10–3 M and [Cu2+(aq)] = 1.0 M into the equation gives

Ecell = 1.10 – log

= 1.10 – log10–3

= 1.10 –× (–3)

= 1.10 + 0.0885

= 1.1885

= 1.2 V

cSubstitute[Zn2+(aq)] = 1.0 M and [Cu2+(aq)] = 10–3 M into the equation

Ecell = 1.10 – log

= 1.10 – log

= 1.10 –× 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)

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

For mixtures where you predict a reaction will occur:

iiwrite separate half equations for the oxidation and reduction reactions

iiiwrite a balanced equation.

aCl2(g) and Zn(s)

bAg+(aq) and Ag(s)

cAg+(aq) and Zn(s)

dZn2+(aq) and Cl–(aq)

A4.

aiReaction will occur.

iiZn(s)→ Zn2+(aq)+2e–(oxidation)

Cl2(g) + 2e–→ 2Cl–(aq) (reduction)

iiiZn(s) + Cl2(g)  Zn2+(aq) + 2Cl–(aq)

bino reaction

ciReaction will occur.

iiZn(s)→ Zn2+(aq)+2e– (oxidation)

2Ag+(aq)+ e–→2Ag(s) (reduction)

iii2Ag+(aq) + Zn(s)  2Ag(s) + Zn2+(aq)

dino 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 1M 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.

acopper(II) sulfate solution is stored in an aluminium container

bsodium chloride solution is stored in a copper container

csilver nitrate solution is stored in a zinc container

dan iron nail is placed in 1 M hydrochloric acid solution

ea plumber uses hydrochloric acid to clean copper pipes

A6.

a3Cu2+(aq) + 2Al(s)  3Cu(s) + 2Al3+(aq)

bno reaction

c2Ag+(aq) + Zn(s)  2Ag(s) + Zn2+(aq)

d2H+(aq) + Fe(s)  H2(g) + Fe2+(aq)

eno reaction

Q7.

Explain the difference between the terms:

aoxidant and reductant

banode and cathode

cconjugate redox pair and conjugate acid–base pair

dexternal circuit and internal circuit

A7.

aOxidant: 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.

banode: electrode at which oxidation occurs

cathode: electrode at which reduction occurs

cconjugate 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.

dexternal 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)

aWhich species is the strongest oxidant and which is the weakest oxidant?

bWhich species is the strongest reductant and which is the weakest reductant?

cLead 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.

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

A8.

aAg+(aq); Mg2+(aq)

bMg(s); Ag(s)

cA 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.

dzinc and magnesium

Q9.

Use the electrochemical series to determine whether:

aelemental iodine is an oxidant or reductant

bcalcium metal is a strong or weak reductant

cthe permanganate ion is a strong or weak oxidant

dnickel is a better reductant than silver

eCu2+(aq) is a better oxidant than Ag+(aq)

fFe2+(aq) can act as an oxidant and a reductant

A9.

aoxidant

bstrong reductant

cstrong oxidant

dtrue

efalse

ftrue

Q10.

aUse the electrochemical series to predict what might be expected to occur if hydrogen gas were bubbled through a solution containing Fe3+ ions.

bWrite an equation for the predicted reaction.

cWhen the reactants were mixed in an experiment, no reaction was observed. Suggest possible reasons for this.

A10.

apredict Fe2+(aq) and H+(aq) are formed

b2Fe3+(aq) + H2(g)  2Fe2+(aq) + 2H+(aq)

cIf 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.

aBromine reacts with iodide ions in solution but does not react with chloride ions.

bHydrogen peroxide (H2O2) solution can spontaneously decompose to form water and oxygen.

cTin metal is added to solutions of tin(II) chloride to prevent oxidation of the tin(II) ions by oxygen in air.

dBlocks of zinc are attached to the iron hulls of ships to reduce corrosion.

A12.

aBromine 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.

bHydrogen 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.

cSn2+ ions can be oxidised to Sn4+ ions by a suitable oxidant. Tin metal can reduce Sn4+ ions so that they re-form Sn2+ ions.

dZinc 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 1M 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.

Heinemann Chemistry 2 (4th edition)