Electrochemistry 1

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(b) How can it be shown that the standard electrode potential for the system

Ga³⁺ + 3e^- Ga

is -0.53 volt?

(c) If the standard electrode potential for the system

Ag⁺ + e^- Ag

is + 0.80 volt, determine the standard emf of a cell with the electrodes in (b) and (c) and give equations for the reactions that take place at each electrode.

Write the equation for the two half cell reactions, the net cell reaction and calculate the cell voltage for each of the following cells:

(a) Pt l H₂(1 atm) l H⁺(1 M) ll Cl^-(1 atm) l Cl₂(g) l Pt

(a) Write the equation for the reaction at the oxidation electrode (anode)

(b) Write the equation for the reaction at the reduction electrode (cathode).

(c) What is the potential difference between the two electrodes before the resistor is connected between them?

(d) What is the potential difference between the two electrodes when chemical equilibrium has been established?

(e) What will be the change in mass of the cadmium rod after 0.0100 mole of electrons have passed through the circuit? Indicate whether this change is an increase or a decrease.

(a) Draw a diagram for an electrochemical cell composed of a standard hydrogen half cell and a half cell in which a silver rod is in contact with 1.00 M silver nitrate solution. Indicate the oxidation electrode, the reduction electrode, the direction of electron flow and the direction of ion migration when the electrodes are connected by a wire.

(b) Write the equation for the spontaneous reaction at the oxidation electrode (anode).

(c) Write the equation for the spontaneous reaction at the reduction electrode (cathode).

(d) What is the potential difference between the two electrodes?

(e) What change occurs in the potential difference when solid sodium chloride is added to the silver nitrate solution?

Write "no change", "increase" or "decrease".

Any one of the halogen molecules or ions might react with iron(II) or iron(III) ions in solution. Using the appropriate standard electrode potentials, write a balanced equation for any reactions that you predict would occur.

Using the above abbreviated table, write an equation for the reaction which occurs in each of the following cases. If there is no reaction, write "no reaction".

(a) 1 M Au³⁺ solution and 1 M Fe²⁺ solution are mixed.

(b) 1 M Au³⁺ solution and 1 M Fe³⁺ solution are mixed.

(c) Au metal is added to 1 M Au³⁺ solution.

(d) Comment on the possibility of preparing a stable 1 M water solution of a gold(I) salt. Give reasons for your answer.

(a) Write the half-cell equation for the reduction electrode reaction.

(b) Write the half-cell equation for the oxidation electrode reaction.

(c) Write the equation for the net cell reaction.

(d) What is the value of the e.m.f. between the Pt and Fe rods?

(a) In the table below are given the constituents of each half cell. In each case predict the net reaction, if any, which will occur. Write the equation for the net reaction, and give the value of the standard potential of the cell, E^o(cell). If you predict no reaction occurs, write "no reaction".

(a) Draw a diagram to illustrate such a cell, and indicate on it all relevant information to show how it works.

(b) What is the potential difference of this cell?

(c) After the cell has operated for a time, the zinc electrode is found to have changed in mass by 0.327 gram. Find the change in the number of mole of

(i) Fe²⁺, (ii) Fe³⁺ and indicate whether the amount is an "increase" or a "decrease".

(a) a cobalt rod immersed in 1 M cobalt(II) nitrate,

(b) a carbon rod immersed in 1 M hydrochloric acid and with chlorine gas at 1 atmosphere pressure being bubbled around the rod.

When the electrodes are connected by a voltmeter it is observed that the cell potential is 1.64 volts, and that the cobalt electrode is negative.

(a) Write the oxidation half equation for the cell reaction

(b) Write the reduction half equation for the cell reaction

(c) Write the equation for the cell reaction

(d) What is the standard electrode (reduction) potential for the cobalt(II)/cobalt couple?

(e) Indicate clearly on the diagram of the functioning cell:

(i) the direction of electron flow

(ii) the directions of +ve ion and -ve ion migration in the bridge

(iii) the oxidation electrode (anode)

(iv) the reduction electrode (cathode)

(f) How would the cell voltage change if the concentration of cobalt(II) nitrate in solution were decreased? (Write "increased", "decreased", or "no change").

1

(a) The standard electrode potential (E^o) of a half-cell system (electrode) represented as

Mⁿ⁺ + ne^- M is the potential of the right hand electrode less that of the left hand electrode in the cell

Pt l H₂(g)(1 at) l H⁺(1 M) ll Mⁿ⁺(1 M) l M

All the components are in their standard states - i.e. solutes have a concentration of 1 M and gases are at a pressure of 1 atmosphere pressure (strictly, 100 kPa).

(b) Connect the standard Ga³⁺ l Ga electrode with a standard hydrogen electrode and observe the potential difference between the two half cells. The Ga electrode will be found to be negative with respect to the standard hydrogen electrode and the potential difference between the two electrodes will be 0.53 volt.

(c) The potential difference for any cell whose components are all in their standard states can be calculated from the tables of standard electrode potentials (also known as standard reduction potentials) as follows:

E^ocell = E^o(larger) - E^o(smaller) = E^o(reduction) - E^o(oxidation).

∴ E^o = (+0.80) - (-0.53) V = +1.33 V

N.B. Be careful with the signs of the component E^o values in the equation. Note that the sign for the cell potential difference calculated must be positive as this means that the reaction is spontaneous.

To determine the reaction direction, the half-cell with the larger E^o value is the one which undergoes reduction while oxidation occurs in the half-cell with the smaller E^o.

Thus as the Ag⁺ l Ag half-cell has the larger E^o value, reduction occurs in the Ag⁺ l Ag half-cell:

Ag⁺ + e^- → Ag

while oxidation occurs in the Ga³⁺ l Ga half-cell:

Ga → Ga³⁺ + 3e^-

Cd²⁺ + 2e^- Cd E^o = -0.40 V

Fe³⁺ + e^- Fe²⁺ E^o = + 0.77 V

As the E^o for the Fe³⁺,Fe²⁺ half-cell is larger, this will be the reduction (cathode) and the oxidation will occur in the Cd²⁺ l Cd half-cell (anode).

(a) Anode reaction: Cd → Cd²⁺ + 2e^-

(b) Cathode reaction: Fe³⁺ + e^- → Fe²⁺

(c) E^o(cell) = E^o(red) - E^o(ox)

= (+0.77) - (-0.40) V

= 1.17 V

(d) After the resistor is connected, electrons can flow from the anode to the cathode via the resistor and the reactions can proceed in both cells. Equilibrium will occur when all of the reactants in each cell have been consumed and then the voltage will be zero.

(e) From the equation above, one mole of electrons passing through the cells causes ½ mole of Cd to be oxidized to Cd²⁺.

∴ 0.0100 mole of electrons would cause 0.00500 mole of Cd to be oxidized.

Mass of cadmium oxidized = 0.00500 x 112.4 g = 0.562 g (decrease in mass).

2H⁺ + 2e^- H₂       0.00 V

Ag⁺ + e^- Ag       +0.80 V

As the E^o for the Ag⁺ l Ag half cell is larger, this half cell will undergo reduction so it is the cathode in the cell while the H⁺ l H₂ half cell is the anode and it undergoes oxidation.

Electrons will flow through the wire from the half-cell where oxidation occurs to the half-cell where reduction occurs, i.e. from the hydrogen half-cell to the silver half-cell.

In the hydrogen half-cell, H₂ molecules are oxidized to H⁺ ions, so there will be excess positive ions in that half cell while in the silver half-cell, Ag⁺ ions are reduced to Ag atoms, creating a deficiency of positive ions there. To maintain electrical neutrality, negative ions move from the bridge to the hydrogen half-cell and positive ions move from the bridge to the silver half-cell.

(b) Anode reaction: H₂ → 2H⁺ + 2e^- (oxidation)

(c) Cathode reaction: Ag⁺ + e^- → Ag (reduction)

(d) E^o(cell) = E^o(red) - E^o(ox)

= (+0.80) - (0.00) = +0.80 V

(e) Addition of Cl^- to the silver half-cell causes silver chloride to precipitate and thus reduces the concentration of Ag⁺ in the silver half-cell. Lowering the [Ag⁺] causes the equilibrium in the cell, represented by the equation

2Ag⁺ + H₂ 2Ag + 2H⁺

to move to the left.

i.e. there is less tendency for the reaction to occur and the cell voltage will be smaller.

As E^o for the F₂ l F^-, Cl₂ l Cl^- and Br₂ l Br^- systems > E^o for Fe³⁺,Fe²⁺ system, then these three halogen systems will undergo reduction and cause the Fe³⁺,Fe²⁺ system to be oxidized.

However, E^o for the I₂ l I^- system is < E^o for the Fe³⁺,Fe²⁺ system so for this combination, the Fe³⁺,Fe²⁺ system will undergo reduction while the I₂ l I^- system will undergo oxidation.

[Note that if the E^o tables are written with the largest values at the top, a reduction (L → R) can be brought about by an oxidation (R → L) below it in the table, but not by an oxidation above it.]

The reaction equations are as follows:

2Fe²⁺ + F₂ → 2Fe³⁺ + 2F^-

2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl^-

2Fe²⁺ + Br₂ → 2Fe³⁺ + 2Br^-

2Fe³⁺ + 2I^- → 2Fe²⁺ + I₂

In each system, the half-cell reaction is

Ag⁺ + e^- Ag

For standard conditions where [Ag⁺] is 1.00 M, E^o = +0.80 V

System 5 has this concentration. Each of the other half-cell systems has a smaller [Ag⁺].

Lowering the [Ag⁺] causes the equilibrium in the half-cell to move to the left.

i.e. there is less tendency for the reaction to occur and the half-cell voltage will be smaller than in half-cell 5. Thus the ranking for the half-cell E values is just the ranking of the concentrations of Ag⁺ in each half-cell.

The solubility product constants for AgCl (half-cell 2) and AgCH₃CO₂ (half-cell 1) have the same units as both salts have the same general formula, AgX and therefore their values can be compared directly. The values are 2 x 10^-10 M² and 3 x 10^-3 M² respectively, so [Ag⁺] is greater in the silver acetate solution. Hence E for system 1 < E for system 2.

In systems 3 and 4 the concentration of Ag⁺ is reduced even further than that in system 2 because ammonia combines with silver ion to form the complex, diamminesilver(I) ion and reduces the free silver ion concentration to very low levels.

Ag⁺ + 2NH₃ [Ag(NH₃)₂]⁺

The larger the [NH₃] in the solution, the smaller will be the [Ag⁺].

Thus the overall ranking of [Ag⁺] and therefore the E values for the half-cell systems is:

3 < 4 < 2 < 1 < 5

(a) A redox reaction between Au³⁺ and Fe²⁺ can occur as E^o for the Fe³⁺,Fe²⁺ system is smaller than E^o for either of the two possible reductions of Au³⁺ listed. Of the two possible reductions of Au³⁺, the reduction to form Au rather than Au⁺ will occur as it has the larger E^o value.

Au³⁺ + 3Fe²⁺ Au + 3Fe³⁺

(b) No reaction can occur as both Au³⁺ and Fe³⁺ are fully oxidized.

For a redox reaction to occur, it requires a reduction half-reaction (left to right in the table) to be combined with an oxidation half-reaction (right to left in the table).

(c) No reaction occurs. The potential redox reaction is the result of the combination of the half-reactions

Au³⁺ + 3e^- → Au (a reduction) and

Au → Au⁺ + e^- (an oxidation)

However, E^o for the overall cell reaction would be

E^o = E^o (reduction) - E^o (oxidation) = (+1.5) - (+1.7)

= - 0.2 V.

A negative E^o(cell) indicates that this reaction is not spontaneous and in fact the reverse reaction is the spontaneous direction.

[When the E^o tables are written with the largest values at the top, a reduction (L → R) can be brought about by an oxidation (R → L) below it in the table, but not by an oxidation above it.]

(d) From the data supplied it can be seen that Au⁺ can be both reduced (to Au) and oxidized (to Au³⁺). The half-reactions are

Au⁺ + e^- → Au (reduction of Au⁺)

Au⁺ → Au³⁺ + 2e^- (oxidation of Au⁺)

and the overall reaction is

3Au⁺ → 2Au + Au³⁺

The E^o(cell) for the reaction = E^o(reduction) - E^o(oxidation)

= (+1.7) - (+1.3) = 0.4 V

Hence a 1 M solution of an Au⁺ salt would decompose to Au metal and an Au³⁺ salt.

Fe³⁺ + e^- Fe²⁺        E^o = + 0.77 V

Fe²⁺ + 2e^- Fe        E^o = - 0.41 V

(a) As E^o for the Fe³⁺,Fe²⁺ system is larger, this half cell will undergo reduction.

Fe³⁺ + e^- → Fe²⁺

(b) As E^o for the Fe²⁺ l Fe system is smaller, this half cell will undergo oxidation.

Fe → Fe²⁺ + 2e^-

(c) 2Fe³⁺ + Fe → 3Fe²⁺

(d) E^o(cell) = E^o(red) - E^o(ox)

= (+0.77) - (-0.41)

= 1.18 V

(a) (i) The two cell half-reactions and their E^o values are

Half-cell A: Fe³⁺ + e^- Fe²⁺ E^o = + 0.77 V

Half-cell B: Sn⁴⁺ + 2e^- Sn²⁺ E^o = +0.15 V

As E^o for half-cell A is larger, this half-cell will undergo reduction and half-cell B will undergo oxidation.

Reduction equation: Fe³⁺ + e^- → Fe²⁺

Oxidation equation: Sn²⁺ → Sn⁴⁺ + 2e^-

Net equation: 2Fe³⁺ + Sn²⁺ → Sn⁴⁺ + 2Fe²⁺

Cell voltage E^o(cell) = E^o(red) - E^o(ox)

= (+0.77) - (+0.15)

= 0.62 V

(ii) The two cell half-reactions and their E^o values are

Half-cell A: MnO₄^- + 8H⁺ + 5e^- Mn²⁺ + 4H₂O E^o = + 1.51 V

Half-cell B: Fe³⁺ + e^- Fe²⁺ E^o = + 0.77 V

As E^o for half-cell A is larger, this half-cell will undergo reduction and half-cell B will undergo oxidation.

Reduction equation: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Oxidation equation: Fe²⁺ → Fe³⁺ + e^-

Net equation: MnO₄^- + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Cell voltage E^o(cell) = E^o(red) - E^o(ox)

= (+1.51) - (+0.77)

= 0.74 V

(iii) The two cell half-reactions and their E^o values are

Half-cell A: UO₂²⁺ + e^- UO₂⁺ E^o = + 0.05 V

Half-cell B: V³⁺ + e^- V²⁺ E^o = - 0.26 V

As E^o for half-cell A is larger, this half-cell will undergo reduction and half-cell B will undergo oxidation.

Reduction equation: UO₂²⁺ + e^- → UO₂⁺

Oxidation equation: V²⁺ → V³⁺ + e^-

Net equation: UO₂²⁺ + V²⁺ → V³⁺ + UO₂⁺

Cell voltage E^o(cell) = E^o(red) - E^o(ox)

= (+0.05) - (-0.26)

= 0.31 V

(b) To retain the ionic charge balance in both cells. This is necessary due to the loss of positively charged ions in half-cell A and the increase of positively charged ions in half-cell B in each case. Ionic charge balance is maintained by K⁺ ions leaving the salt bridge into half-cell A and Cl^- ions leaving the salt bridge into half-cell B. Note that ions do not migrate from half-cell A to half-cell B - if this were to occur, the redox reaction could take place directly as the half-cells would no longer be isolated and no electrons would flow through the wire.