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Name/CG: _________________________________ JC2 H2 Chemistry Electrochemistry Revision Duration 25 min 1(a) The following data refer to the ions of the element vanadium in acidic solution.

Ions Colour of aqueous ions

VO2+ Yellow

VO2+ Blue V3+ Green V2+ Violet

For each of the following situations, use Eo

cell data to predict what might happen when the two reagents are mixed together. Write balanced equations for any reactions that occur.

(i) excess Zn(s) and VO2+ (aq)

Starting reagent is VO2+ and Zn To check if Zn can reduce VO2+.

VO2+ + 2H+ + e- V3+ + H2O E θ = +0.34V Zn2+ + 2e- . Zn E θ = -0.76V Eθcell = Eθred - Eθoxid

= +0.34 – (-0.76) = +1.10V [1/2] Since Eθcell > 0, reaction is feasible.

Zn(s) + 2VO2+ (aq) + 4H+(aq) → Zn2+(aq) + 2V3+ (aq) + 2H2O (l) [1/2] Colour change of solution from blue to green. [1/2]

To check if Zn can reduce V3+.

V3+ + e- . V2+ E θ = -0.26V Zn2+ + 2e- . Zn E θ = -0.76V

Eθcell = Eθred - Eθoxid

= -0.26 – (-0.76) = +0.50V [1/2] Since Eθcell > 0, reaction is feasible.

Zn(s) + 2V3+ (aq) → Zn2+(aq) + 2V2+ (aq) [1/2]

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Colour change of solution from green to violet. [1/2] In conclusion,

Zn(s) + VO2+ (aq) + 2H+(aq) → Zn2+(aq) + V2+ (aq) + H2O (l) Zn can reduce VO2+ to V2+. Solution changes from blue to green and finally violet in colour. (ii) S4O6

2- (aq) and V2+(aq) [4]

S4O6

2-+ 2e- 2S2O32- +0.09 (Reduction)

V3+ + e- . V2+ -0.26 (Oxidation)

Eθcell = Eθred - Eθoxid

= 0.09 – (-0.26) = +0.35V [1/2]

Since Eθcell > 0, reaction is feasible.

Solution changes from violet to green. [1/2]

S4O62- (aq) + 2V2+ (aq) → 2S2O3

2- (aq) + 2V3+ (aq) [1/2]

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(b) The magnitudes (but not the sign) of the standard electrode potentials of two metals B and C are given below:

Equation Magnitude of Eo

B2+ + 2e B

0.30

C2+ + 2e C

0.36

When B2+/B is connected to the standard hydrogen electrode, electrons flow from the B electrode to the platinum electrode of the standard hydrogen electrode. When the half-cells of B2+/B and C2+/C are connected, electrons flow from the B electrode to the C electrode.

(i) Deduce the signs of the Eo values of the B2+/B and C2+/C half-cells. Explain

your answer.

Since electrons flow from the B electrode (i.e anode/oxidation) [1/2] to the platinum electrode of SHE and Eθcell = Eθred - Eθoxid> 0

OR ⇒ 0.00 – (– 0.30) > 0 [1/2] ⇒ sign of the Eθ value of the B2+/B is negative [1/2]

Since electrons flow from the B electrode (i.e anode/oxidation) to the C electrode and Eθcell = Eθred - Eθoxid> 0

⇒ +0.36 – (– 0.30) > 0 [1/2] ⇒ sign of the Eθ value of the C2+/C is positive [1/2]

or other logical reasoning / working (ii) Calculate the e.m.f. of the cell made up of B2+/B and C2+/C half-cells.

emf of cell = +0.36 – (– 0.30) = + 0.66 V [1]

no units / + sign – minus ½ (iii) What would be the effect on the e.m.f. of the cell in (b)(ii) when the

concentration of C2+ is decreased.

C2+ + 2e C

When [C2+] decreased, • the equilibrium position shifts to the left [1/2] to increase [C2+] by Le

Chatelier’s Principle. no mark if eqm is not written • Oxidation is more likely to occur. • Electrode potential Eθ decreases [1/2]

Hence Eθcell = Eθ - Eθ decreases [1].

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(iv) Suggest a replacement half-cell for B2+/B half-cell which would reverse the direction of electron flow in the C2+/C half-cell.

[8] Eθcell = Eθred - 0.36 > 0 Choose half-cell with Eθ> 0.36V

Ag+/ Ag [1] or other correct half-cell

(c) Explain the following observations as fully as you can.

The electrolysis of dilute potassium bromide chloride produces oxygen gas at the anode, whereas the electrolysis of concentrated potassium bromide chloride results in the formation of a orange-red a yellowish-green colouration at the anode region.

[4]

Cl2 + 2e 2Cl- E = +1.36 V 4H+ (l) + O2 (g) + 4e 2 H2O (l) E = +1.23 V

For dilute potassium bromide: Both Cl- and H2O migrate to anode

H2O is preferentially discharged as it has a less positive Eθ [1/2] value compared to Cl- Oxidation: 2 H2O (l) → 4H+ (l) + O2 (g) + 4e [1/2]

Hence effervescence of O2 gas is observed

For concentrated potassium bromide: Both Cl- and H2O migrate to anode

Cl- is preferentially discharged even though it is has a more positive Eθ value compared to H2O because the concentration of Cl- is very high. [1/2] Oxidation: 2Cl- (aq) → Cl2 (aq) + 2e [1/2] Hence effervescence of pale yellow due to Cl2 is observed.

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RJC Prelim 05/P2/Q2 (18 min) 2 Aluminium/air batteries are used as back-up power supplies in many telephone

exchanges. The structure of one such battery is shown below: The electrolyte used is aqueous potassium hydroxide. In the reaction, oxygen in

air is reduced while aluminium is oxidised as shown by the following equation:

Al (s) + 4OH−(aq) ⎯→ Al(OH)4−(aq) + 3e− (a) Is terminal Y negative or positive? Give a reason for your answer. [2] Terminal Y is negative. This is because electrons are produced by the oxidation of aluminium and leave

the battery via terminal Y. (b) Write a balanced equation with state symbols for the reaction that takes place at

the porous electrode Z. [1] O2

(g) + 2H2O (l) + 4e− ⎯→ 4OH− (aq)

(c) Why must electrode Z be porous? [1]

Z must be porous so that oxygen may diffuse through the electrode and come into contact with the electrolyte where it is reduced.

(d) Sometimes, electrode Z becomes clogged up so that it is no longer porous. This causes the aluminium/air battery to be potentially explosive. With the help of a suitable equation, explain why this is so. [2]

Instead of oxygen being reduced, water may be reduced at electrode Z forming

hydrogen gas: 2H2O (l) + 2e− ⎯→ H2 (g) + 2OH− (aq) H2 gas is produced in a sealed container, so potentially explosive.

(e) The aluminium used is specially alloyed so that it is not easily oxidised by air. Explain why ordinary aluminium cannot be used. [1]

aluminium alloy

porous electrode Z

terminal Y

terminal X

electrolyte

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Ordinary aluminium is protected by a dense, impervious layer of aluminium oxide. This layer prevents aluminium from being oxidised so that it cannot act as an anode.

(f) Besides aluminium, zinc may also be used. Give two advantages of using

aluminium rather than zinc. [2]

Zn2+ + 3e– Zn EO = – 0.76V Al3+ + 3e– Al EO = – 1.66V Aluminium has a more negative standard reduction potential/ bigger standard

oxidation potential than zinc so that the operating voltage of aluminium/air battery is higher than zinc/ air battery.

Aluminium produces more electron per mole of metal so that more electrons can be produced from aluminium/air battery than zinc/ air battery for the same mass of metal used.

Aluminium is lighter than zinc (Accept other sensible answers, but not aluminium is cheaper than Zn, less

corrosive than Zn)

(g) When aqueous potassium hydroxide is replaced by aqueous sodium chloride, it is noticed that a white solid is formed around the aluminium alloy electrode. Name the white solid and explain why it is formed. [2]

The white solid is aluminium hydroxide. There are insufficient hydroxide ions to dissolve the insoluble amphoteric

aluminium hydroxide. Al(OH)3 is formed due to the precipitation reaction between Al3+ (produced by oxidation of Al) and OH− (produced by reduction of oxygen.)

(h) The aluminium/air battery with aqueous potassium hydroxide as an electrolyte

was used as an electrical source to electroplate an article with copper. If the mass of aluminium electrode decreased by 5.4 g, calculate the mass of copper that had plated out of the solution. [2]

Al ≡ 3e− Cu2+ (aq) + 2e− ⎯→ Cu (s) Hence 2 Al ≡ 3 Cu No.of mol of Al used = 5.4/27 = 0.2 No of mol of Cu formed = 3/2 x 0.2 = 0.300 Mass of Cu = 0.300 x 63.5 g = 19.1 g

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NJC Prelim 05/P2/Q2 (15 min) 3(a) Iron is the most widely used d-block metal in the construction industry, mainly

because it is very abundant in the earth’s crust. However, we are also familiar with the red-brown solid known as rust, which weakens the structural strength of iron. The control of corrosion of iron is hence of great economic importance.

The scheme below outlines the stages in the rusting process.

(i) Why do you think that the structural strength of iron is weakened by the

formation of rust?

Formation of rust, an ionic compound, will weaken the structural strength of iron, as rust is brittle.

(ii) Stage I of rusting is an electrochemical process.

Taking A to be the anode, and C to be the cathode of the electrochemical cell reaction of rusting in the above diagram,

1 write ion-electron equations for the process at

A : Fe (s) → Fe2+ (aq) + 2e─

C : O2 + 2H2O + 4e─ → 4OH─

2 By means of an arrow, show the direction of electron flow in the above cell.

3 Show, by a cross (X) on the diagram, one spot where you would expect most rust to be accumulated.

moist air

Stage I

prolonged exposure to air

Fe(s) Fe(OH)2(s) Fe2O3.xH2O(s) red-brown rust

A

C

air

iron

water

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(iii) Use information from your Data Booklet, estimate a value for the potential

of the cell in (a)(ii), giving reasons.

Ecello = 0.44 – (- 0.40) = +0.84V

The value of the potential of the cell in (a)(ii) will be less than +0.84V as the conditions of the cell is not of standard conditions.

[6]

(b) Unlike iron, aluminium has resistance to atmospheric corrosion because of the oxide layer on its surface. (i) State one common use of aluminium which makes use of this property.

Cans for soft drinks, car bodies. (ii) This oxide layer can be thickened by an electrolytic process.

On the space below, draw a fully-labelled diagram to show how this process can take place.

[4]

Aluminium Anode

dil H2SO4 (aq)

Graphite Cathode

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Homework Duration: 9 min 1 Predict the products formed at the anode, and at the cathode, when the following

liquids are electrolyzed using inert electrodes. (a) NaBr(l) For electrolysis: At anode(+): Br- is oxidized to Br2. Hence product is Br2. At cathode(-): Na+ is reduced to Na. Hence product is Na. (b) NaBr(aq) Species present in dilute sodium chloride: Na+, Br-, H2O

At the anode (+) • Both H2O and Br- migrate to anode • Preferential oxidation of Br- as EӨ of Br2 / Br- (+1.07 V) is less positive than EӨ of

O2 / H2O (+1.23V) • Hence product is Br2

At the cathode (-)

• Both H2O and Na+ migrate to cathode • Preferential reduction of H2O as EӨ of H2O / H2 (-0.83 V) is more positive than EӨ

of Na+ / Na (-2.71V) • Hence product is H2

(c) CuF2(aq) Species present in dilute sodium chloride: Cu2+, F-, H2O

At the anode (+) • Both H2O and F- migrate to anode • Preferential oxidation of H2O as EӨ of O2 / H2O (+1.23V) is less positive than EӨ

of F2 / F-- (+2.87 V) • Hence product is O2

At the cathode (-)

• Both H2O and Cu2+ migrate to cathode • Preferential reduction of Cu2+ as EӨ of Cu2+ / Cu (+0.34 V) is more positive than

EӨ of H2O / H2 (-0.83V) • Hence product is Cu

[6]

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Duration: 6 min 2(a) By means of a fully labelled diagram, describe how the standard electrode

potential of Cr3+/Cr2+ system can be measured by using standard hydrogen electrode.

[4]

[diagram [2], anything missing or wrong -1/2 till zero] (b) Use the Data Booklet to predict the outcome of mixing acidified aqueous hydrogen peroxide with (i) aqueous bromine, Br2 is reduced to Br- H2O2 is oxidised to O2 Eθcell = 1.07 – 0.68 = +0.39 V [1/2] > 0 and reaction is feasible Br2(aq) + H2O2 → 2Br-(aq) + O2(g) + 2H+(aq) [1] no state symbol – minus ½

Reddish-brown aqueous bromine is decolourised [1/2] and effervescence of O2 gas is seen [1/2].

(ii) aqueous iron(II) sulfate. Include Eo

cell, balanced equation and observation in your answer. Fe2+ is oxidised to Fe3+ H2O2 is reduced to H2O Eθcell = 1.77 – 0.77 = +1.00 V [1/2] > 0 and reaction is feasible 2Fe2+ (aq) + H2O2 + 2H+(aq) → 2Fe3+(aq) + 2H2O(l) [1] no s.s. – minus ½ Green Fe2+ (aq) solution turns to yellow Fe3+(aq) [1/2].

[4]

Voltmeter

Cr2+ ions (1 mol dm-3) & Cr3+ ions (1 mol dm-3)

Pt (s)

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Duration: 15 min 3(a) An electrochemical cell has an overall e.m.f. of +1.33 V at 25°C. The overall cell

reaction is shown below.

2CO (g) + O2 (g) → 2CO2 (g) The cell diagram is given as:

Ni (s) | CO (g), CO2 (g) | CO32- (aq) || CO3

2- (aq) | O2 (g), CO2 (g) | Ag (s) The reaction occurring at the silver electrode is:

2CO2 (g) + O2 (g) + 4e- → 2CO32- (aq) Eθ = +0.69 V

(i) Deduce the reaction occurring at the nickel electrode and calculate the

standard electrode potential for the reaction involved. Anode: 2CO (g) + 2CO3

2- (aq) → 4CO2 (g) + 4e- [1] Eθcell

= Eθred – Eθoxid +1.33 = +0.69 – Eθoxid Eθoxid

= 0.69 – 1.33 = –0.64 V [1]

(ii) Draw a labelled diagram to show the laboratory experimental set-up of the

electrochemical cell under standard conditions. Indicate the direction of the electron flow in your sketch.

[3m, anything missing – minus ½]

V

Ag electrode

Ni electrode

[CO32- (aq)] = 1moldm-3 [CO3

2- (aq)] = 1moldm-3

CO2 (g) and CO (g) at 298K and 1atm

CO2 (g) and O2 (g) at 298K and 1atm salt bridge

Electron flow

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(iii) Suggest a replacement half cell, involving a gas, for CO32- (aq) | O2 (g),

CO2 (g) | Ag (s), which would reverse the direction of the electron flow in the CO3

2- (aq) | CO2 (g), CO (g) | Ni(s) half cell. Your answer needs to state both the electrode and the reagents of your new half-cell.

[6] To reverse electron flow, the selected half cell should now undergo oxidation & the other half cell with the Ni electrode will undergo reduction (Eθred

= –0.64 V) Eθcell

= Eθred – Eθoxid Eθcell

= –0.64 – Eθoxid > 0 ⇒ Eθoxid < –0.64 V

Pt (s) | H2O (l) | OH- (aq) | H2 (g) [2] Or Pt(s) [1]

KOH(aq), H2(g) [1] (b) A modified version of the Leclanche cell is now widely used as a dry cell battery

to power portable gadgets.

Two possible half-cell reactions in the dry cell are:

[Zn(NH3)4]2+ (aq) + 2e- Zn (s) + 4NH3 (aq) Eθ = -1.03 V NH4

+ (aq) + MnO2 (s) + H2O (l) + e- Mn(OH)3 (s) + NH3 (aq) Eθ = +1.00 V (i) Calculate the standard electrode potential, Eθ of the dry cell.

Eθcell = Eθred – Eθoxid

= 1.00 – (-1.03) = +2.03 V [1]

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(ii) The actual e.m.f. of the cell is +1.05 V. Give a reason as to why this value

is different from the one calculated in (b)(i). Due to non-standard conditions / non-standard concentrations. [1]

(iii) Determine the amount of time in seconds required for a dry cell containing

6.50 g of MnO2 to produce a constant current of 2.0 × 10−3 A.

[4] No. of mol of MnO2 = 6.50 / 86.9 = 0.0748 mol [1]

1 mol of MnO2 ≡ 1 mol of e-

0.0748 =

t = 3.61 × 106 s [1]