CHEM1901/3 2014-J-10 June 2014

• At a temperature of absolute zero, the entropy of deuterated methane CH3D is 12 J K–1 mol–1. Explain the significance of this value and suggest an explanation for it.

Marks 2

The 3rd Law of Thermodynamics is sometimes written in the form below:

• The entropy of a perfect crystal at absolute zero is exactly equal to zero. CH3D will adopt the same crystal structure as CH4 as there is no difference in its shape or size. The relative position of the D atom in neighbouring molecules, however, will be completely random as there will minimal energy difference between them.

A crystal of CH3D will not be perfect even at absolute zero.

CHEM1901/3 2014-J-12 June 2014

A voltaic cell consists of Cd2+/Cd and Ag+/Ag half cells with initial concentrations of [Cd2+] = 1.00 M and [Ag+] = 0.60 M. Each half cell contains 1.00 L of solution.

Marks 4

What is the voltage of the cell at 20 °C after equilibrium has been reached? 0 V

What are the concentrations of the Cd2+(aq) and the Ag+(aq) ions at 20 °C after equilibrium has been reached?

The standard reduction half-cell reactions are (from the data sheet):

Cd2+(aq) + 2e– à Cd(s) E° = -0.40 V Ag+(aq) + e– à Ag(s) E° = +0.80 V

As the Cd2+ / Cd value is the least positive, it is reversed and the reaction and cell potential are:

2Ag+(aq) + Cd(s) à 2Ag(s) + Cd2+(aq) E° = (+0.40 V) + (+0.80 V) = +1.20 V

Using E° = 𝑹𝑻𝒏𝑭

lnK, the equilibrium constant for this 2 e- process at T = 20 °C is:

K = exp𝟏.𝟐𝟎  𝐕  ×  𝟐  ×  (𝟗𝟔𝟒𝟖𝟓  𝐂  𝐦𝐨𝐥!𝟏)𝟖.𝟑𝟏𝟒  𝐉  𝐊!𝟏  𝐦𝐨𝐥!𝟏  ×  (𝟐𝟗𝟑  𝐊)

= 2.03 × 1044

The equilibrium constant is so large that the reaction essentially goes to completion. Initially, [Cd2+(aq)]initial = 1.00 M and [Ag+(aq)] initial = 0.60 M. If essentially all of the Ag+(aq) reacts then [Cd2+(aq)] = 1.30 M. If the small amount of Ag+(aq) left over has [Ag+(aq)] = x M:

Ecell = E° - 𝑹𝑻𝒏𝑭

lnQ = E° - 𝑹𝑻𝟐𝑭

ln[𝐂𝐝𝟐! 𝐚𝐪 ][𝐀𝐠! 𝐚𝐪 ]𝟐

= 0

= (1.20 V) - 𝟖.𝟑𝟏𝟒  𝐉  𝐊!𝟏  𝐦𝐨𝐥!𝟏  ×  (𝟐𝟗𝟑  𝐊)

𝟐  ×  (𝟗𝟔𝟒𝟖𝟓  𝐂  𝐦𝐨𝐥!𝟏)  ln (𝟎.𝟏𝟑)[𝐀𝐠! 𝐚𝐪 ]𝟐

This gives [Ag+(aq)] = 2.6 × 10-21

[Cd2+]eq = 1.3 M [Ag+]eq = 2.6 × 10-21 M

CHEM1901 2012-J-11 June 2012

• A voltaic cell is set up at 298 K based on the following half-cell reactions.

Pt2+(aq) + 2e– Pt(s) E° = +1.18 V

Rh3+(aq) + 3e– Rh(s) E° = +0.76 V

Write the overall chemical reaction that takes place in this cell.

Marks7

3Pt2+(aq) + 2Rh(s) → 3Pt(s) + 2Rh3+(aq)

Write the same reaction in shorthand voltaic cell notation.

Rh(s) | Rh3+(aq) | | Pt2+(aq) | Pt(s)

Which metal electrode is acting as the cathode in this reaction? Pt

Calculate the potential of this cell at 298 K, in which the concentration of Pt2+(aq) is 0.0544 mmol L–1 and the concentration of Rh3+(aq) is 0.0393 mmol L–1.

As the Rh3+ / Rh reduction potential is the smallest, it is reversed and becomes the oxidation half cell (as in the chemical reaction above). Hence Eo

= Eoreduction + Eo

oxidation = (+1.18 V) + (-0.76 V) = +0.42 V At the non-standard concentrations, the Nernst equation can be used to calculate the cell potential for this 6 electron reaction

Ecell = E° - 𝑹𝑻𝒏𝑭

lnQ = E° - 𝑹𝑻𝟔𝑭

ln[𝐑𝐡𝟑!(𝐚𝐪)]𝟐

[𝐏𝐭𝟐!(𝐚𝐪)]𝟑

= (+0.42 V) - 𝟖.𝟑𝟏𝟒  𝐉  𝐊!𝟏  𝐦𝐨𝐥!𝟏 (𝟐𝟗𝟖  𝐊)

𝟔 (𝟗𝟔𝟒𝟖𝟓  𝐂  𝐦𝐨𝐥!𝟏) ln(𝟎.𝟎𝟑𝟗𝟑  ×  𝟏𝟎!𝟑)𝟐

(𝟎.𝟎𝟓𝟒𝟒  ×  𝟏𝟎!𝟑)𝟑 = +0.38 V

Answer: +0.38 V

CHEM1901/3 2010-J-12 June 2010

• A voltaic cell is set up at 298 K based on the following reaction

Cr3+(aq) + Al(s) Cr(s) + Al3+(aq)

Express the overall reaction in voltaic cell notation.

Marks5

Al(s) |||| Al3+(aq) |||||||| Cr3+(aq) |||| Cr(s)

Calculate the cell potential at 298 K when the concentration of Cr3+(aq) is 0.213 M and the concentration of Al3+(aq) is 0.078 M.

The two half cell reactions and standard potentials are: Cr3+(aq) + 3e- ���� Cr(s) E°reduction = -0.74 V

Al ���� Al3+(aq) + 3e E°oxidation = +1.68 V

Hence, the standard cell potential is: E° = E°oxidation + E°reduction = (+1.68 – 0.74) V = +0.94 V

At the non-standard concentrations, the Nernst equation can be used to work out the potential for this 3-electron process:

Ecell = E° - ��

�� lnQ

= (+0.94 V) - ��.� � � ��������� �

�������� � ����� ln

��.����

��.��� = +0.95 V

Answer: +0.95 V

Calculate the equilibrium constant at 298 K.

Using E° = ��

�� lnK,

+0.94 V = ��.� � � ��������� �

�������� � ����� lnK

K = 4.9 × 1047

Answer: 4.9 × 1047

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2010-J-12 June 2010

• How long (in seconds) would it take for all the gold to be plated out of 55.0 mL of a 2.34 × 10–3 mol L–1 solution of Au3+(aq), using a current of 0.75 A?

2

The number of moles of Au3+(aq) to be reduced is: number of moles of Au3+ = concentration × volume = (2.34 × 10-3 mol L-1) × (0.0550 L) = 0.000129 mol The reduction of each Au3+ ion requires 3 electrons so the number of moles of electrons required is: number of moles of electrons = 3 × 0.000129 mol = 0.000386 mol

A current I passing for a time t delivers It / F moles of electrons. Hence: t = number of moles of electrons × (F / I ) = (0.000386 mol) × (96485 C mol-1) / (0.75 C s-1) = 50. s

Answer: 50. s

CHEM1901/3 2008-J-13 June 2008

• A voltaic cell consists of Zn2+/Zn and Cu2+/Cu half cells with initial concentrations of [Zn2+] = 1.00 M and [Cu2+] = 0.50 M. Each half cell contains 1.00 L of solution.

Marks4

What is the voltage of the cell at 20 °C after equilibrium has been reached?

Ecell = 0 V

What are the concentrations of the Ni2+(aq) and the Cu2+(aq) ions at 20 °C after equilibrium has been reached?

The standard reduction half-cell reactions are (from the data sheet):

Cu2+(aq) + 2e– ���� Cu(s) E° = +0.34 V Zn2+(aq) + 2e– ���� Zn(s) E° = -0.76 V

As the Zn2+ / Zn value is the least positive, it is reversed and the reaction and cell potential are:

Cu2+(aq) + Zn(s) ���� Cu(s) + Zn2+(aq) E° = (+0.34 V) + (+0.76 V) = +1.10 V

Using E° = ���� lnK, the equilibrium constant for this 2 e- process at T = 20 °C (=

20 + 273 K) is:

K = exp���.� �� � ������ � ��������.��� � ��� ������ ���� �� � = 6.97 × 1037

The equilibrium constant is so large that the reaction essentially goes to completion. As, initially, [Zn2+(aq)]initial = 1.00 M and [Cu2+(aq)] initial = 0.50 M, the latter is the limiting reagent and so:

[Zn 2+(aq)]equilibrium = 1.50 M and [Cu2+(aq)] = 0.00 M

Alternatively, a reaction table can be constructed:

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

initial 0.50 1.00

change -x +x

final 0.50 - x 1.00 + x Hence:

K = � !�"�#$�%��&�"�#$�% =

�.'(.�)( = 6.97 × 1037

As K is very large, x ~ 0.5 M.

[Zn2+]eq = 1.5 M [Cu2+]eq = 0.0 M

CHEM1901/3 2008-J-14 June 2008

• Judging by the electrochemical series of standard reduction potentials, H+(aq) should be reduced to H2(g) by exposure to metallic Fe, which is oxidised to Fe2+(aq). Why do we not see water on iron spontaneously generating hydrogen gas?

Marks 2

The standard electrode potentials correspond to reactants having standard concentrations – they refer to [H+(aq)] = 1.0 M. Water has a pH close to 7 and so [H+(aq)] = 10-7 M. Also, there is a high overpotential associated with the formation of H2(g) and so the standard electrode potential does not give the actual potential required to produce H2(g).

CHEM1901/3 2006-J-10 June 2006

• Refer to the electrochemical potentials on the data sheet. Marks

3 Show that Fe2+(aq) is not stable under 1 atm O2 in a 1 M solution of HCl. What happens to Fe2+?

Fe2+ will be oxidised by O2 under these conditions. The relevant half cells are:

Fe2+(aq) ���� Fe3+(aq) + e-…………….E° = −−−−0.77 V O2(g) + 4H+(aq) + 4e- ���� 2H2O E° = +1.23 V

The overall reaction is:

4Fe2+(aq) + O2(g) + 4H+(aq) ���� 4Fe3+(aq) + 2H2O E° = −−−−0.77 + 1.23 = +0.46

As E° > 0, the reaction is spontaneous.

Show Fe(s) is stabilised by galvanizing with Zn(s).

The electrode potentials for oxidation of Fe(s) and Zn(s) are:

Fe(s) ���� Fe2+(aq) + 2e-…………….E° = ++++0.44 V

Zn(s) ���� Zn2+(aq) + 2e-…………….E° = ++++0.76 V

As Zn(s) has a higher oxidation potential, it will be preferentially oxidised and will therefore protect Fe(s).

Show Cu+(aq) is not stable in water. What would happen to Cu+(aq)?

The electrode potential for reduction of Cu+(aq) is:

Cu+(aq) + e- ���� Cu(s)… ………….E° = ++++0.53 V

The electrode potential for oxidation of Cu(s) is:

Cu(s) ���� Cu2+(aq) + 2e-…………….E° = −−−−0.34 V

The reaction:

2Cu+(aq) ���� Cu(s) + Cu2+(aq)

therefore has an overall potential of E° = (+0.53) + (−−−−0.34) = ++++0.17 V

This disproportionation reaction has E° > 0 and is spontaneous.

CHEM1901/3 2004-J-10 June 2004

Consider a solution that is 1 M in both H+ and Fe

2+ ions under 1 atm of oxygen gas.

Use the standard reduction potentials on the data sheet to show that Fe2+

ions are

unstable under these conditions.

Marks

6

Using the half cells:

Fe2+

(aq) Fe3+

(aq) + e– E°ox = –0.77 V

O2(g) + 4H+(aq) + 4e

– 2H2O(l) E°red = +1.23 V

The overall reaction is:

O2 + 4H+(aq) + 4Fe

2+(aq) 2H2O(l) + 4Fe

3+(aq) E°cell = +0.46 V

As E°cell > 0, the reaction is spontaneous.

Why can such solutions be preserved by the addition of iron metal?

When Fe(s) and Fe2+

(aq) are both present, the oxidation half cells:

Fe2+

(aq) Fe3+

(aq) + e– E°ox = –0.77 V

Fe(s) Fe2+

(aq) + 2e– E°ox = +0.44 V

are competing. As oxidation of Fe(s) has a more positive oxidation potential,

Fe(s) is oxidised preferentially to Fe2+

(aq). The Fe(s) is present in large excess as

a piece of solid metal is added so very little Fe2+

(aq) is oxidized.

Is the following reaction spontaneous?

Fe(s) + 2Fe3+

(aq) 3Fe2+

(aq)

Using the half cells:

Fe3+

(aq) +e- Fe

2+(aq) E°red = +0.77 V

Fe(s) Fe2+

(aq) + 2e– E°ox = +0.44 V

The overall cell potential is E°cell = {(+0.77) + (+0.46)} V = 1.23 V

As E°cell > 0, the reaction is spontaneous.

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1901/3 2004-J-10 June 2004

Calculate the reduction potential of a half-cell consisting of a platinum electrode

immersed in a 2.0 M Fe2+

and 0.20 M Fe3+

solution at 25 C.

The reduction potential for Fe3+

(aq) + e- Fe

2+(aq) has E°cell = 0.77. This refers

to standard concentration. Using the Nernst equation, the electrode potential at

the given concentrations becomes:

Ecell = E° - 2.303 × 𝑹𝑻

𝒏𝑭logQ = E° - 2.303 ×

𝑹𝑻

𝒏𝑭log

[𝐅𝐞𝟐+ 𝐚𝐪 ]

[𝐅𝐞𝟑+ 𝐚𝐪 ] =

= (+0.77 V) - 2.303 × 𝟖.𝟑𝟏𝟒 𝐉 𝐊−𝟏𝐦𝐨𝐥−𝟏 (𝟐𝟗𝟖 𝐊)

𝟐 ×(𝟗𝟔𝟒𝟖𝟓 𝐂 𝐦𝐨𝐥−𝟏) log

𝟐.𝟎

𝟎.𝟐

= +0.71 V

Answer: 0.71 V