Electrochemistry [Ans]

8/19/2019 Electrochemistry [Ans]

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Tutorial 20: Electrochemistry – Suggested Solutions

1(a)(1)

(i) Cr 2O72 – (aq) + 14H+ (aq) + 6e ⇌ 2Cr 3+ (aq) + 7H2O (l ) E = +1.33 V

Cu2+(aq) + 2e- ⇌ Cu (s) E = +0.34 V

Cathode: Pt electrode in Cr 2O72 – (aq) / Cr 3+ (aq) half –cell

Red: Cr 2O72 – (aq) + 14H+ (aq) + 6e 2Cr 3+ (aq) + 7H2O (l ) [1]

Anode: Copper electrode in Cu2+ (aq) / Cu (s) half –cell

Oxid: Cu (s) Cu2+ (aq) + 2e [1]

(ii) Cr 2O72 – (aq) + 14H+(aq) + 3Cu(s) 2Cr 3+ (aq) + 7H2O(l ) + 3Cu

2+(aq) [1]

(iii) Ecell = +1.33 – (+0.34)

= +0.99 V [1]

(iv)

1(a)(2)

(i) H2O2 (aq) + 2H+ (aq) + 2e ⇌ 2H2O (l ) E = +1.77 V

O2 (g) + 2H+ (aq) + 2e ⇌ H2O2 (aq) E = +0.68 V

Cathode: Pt electrode in H2O2 (aq) / H2O (l ) half –cell

Red: H2O2 (aq) + 2H+ (aq) + 2e 2H2O (l ) [1]

Anode: Pt electrode in O2 (g) / H2O2 (aq) half –cell

Oxid: H2O2 (aq) O2 (g) + 2H

+ (aq) + 2e [1]

(ii) 2H2O2 (aq) 2H2O (l ) + O2 (g) [1]

(iii) Ecell = +1.77 – (+0.68)

= +1.09 V [1]

(iv)

V

T = 298 K

Pt

Cusalt bridge

[Cu +(aq)]= 1 mol dm –3

[Cr 2O7 – (aq)]

= [Cr 3+(aq)]= [H+(aq)]= 1 mol dm –3

e – Correct drawing & labeling [1]

Correct conditions [1]Correct e – direction [1]

V

T = 298 K

Pt[H2O2(aq)]= [H+(aq)]= 1 mol dm –3

Pt

O2 (1 atm)

[H2O2(aq)]= [H+(aq)]= 1 mol dm –3

salt bridge

e – Correct drawing & labeling [1] Correct conditions [1]Correct e – direction [1]


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1(b)(i) CO32- precipitates out Cu2+ so that [Cu2+(aq)] decreases. [1]

This shifts the position of equilibrium in Cu2+(aq) + 2e ⇌ Cu(s) to the left so that EL decreases. Hence Ecell = ER – EL increases (i.e. Ecell > Ecell). [1]

(ii) Br – reduces Cr 2O72 to Cr 3+.

This decreases [Cr 2O72 (aq)] but increases [Cr 3+ (aq)] [1] so that the position of

equilibrium in Cr 2O72 –(aq) + 14H+(aq) + 6e ⇌ 2Cr 3+(aq) + 7H2O(l ) shifts to the left and so

ER decreases.Hence Ecell = ER – EL decreases (i.e. Ecell < Ecell). [1]

1(c)(i) To decrease Ecell, either decrease ER or increase EL.

(1) ER can be decreased by using H2O2 or H+ of a lower concentration in the H2O2/H2O

half –cell. [1]

(2) EL can be increased by using O2 of a higher pressure, H+ of a higher concentration

or H2O2 of a lower concentration in the O2 / H2O2 half –cell. [1]

(ii) Hydrogen peroxide can spontaneously undergo oxidation and reduction under standardconditions:

2H2O2 (aq) 2H2O(l ) + O2 (g), Ecell = +1.77 – (+0.68)V = +1.09V > O [1]

reaction is energetically feasible and spontaneous under standard conditions.

Hence bubbles of oxygen are given off by a solution of hydrogen peroxide on standing.

Disproportionation. [1] (In general, substances disproportionate if their ER > EL.)

2(a) I2 (aq) + 2e ⇌ 2I (aq) E = +0.54 V

Fe2+ (aq) + 2e ⇌ Fe (s) E = –0.44 V

Ecell = –0.44 – (+0.54)

= –0.98 V < 0 [1]

reaction is not likely to occur.

(b) H2O2 (aq) + 2H+ (aq) + 2e ⇌ 2H2O (l ) E = +1.77 V

I2 (aq) + 2e ⇌ 2I (aq) E = +0.54 V

Ecell = +1.77 – (+0.54)

= +1.23 V > 0 [1]


H2O2 (aq) + 2H+ (aq) + 2I (aq) 2H2O (l ) + I2 (aq) [1]

A brown solution is formed. [1]

(c) Cr 2O72 – (aq) + 14H+ (aq) + 6e ⇌ 2Cr 3+ (aq) + 7H2O (l ) E = +1.33 V

SO42 – (aq) + 4H+ (aq) + 2e- ⇌ SO2 (aq) + 2H2O (l ) E = +0.17 V

Ecell = +1.33 – (+0.17)

= +1.16 V > 0 [1]


Cr 2O72 –(aq) + 2H+ (aq) + 3SO2 (aq) 2Cr

3+ (aq) + H2O (l) + 3SO42 – (aq) [1]

Orange solution turns green. [1]


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2(d) O2 (g) + 2H2O (l ) + 4e

⇌ 4OH (aq) E = +0.40 V

Cr 3+ (aq) + e ⇌ Cr 2+ (aq) E = –0.41 V

Ecell = +0.40 – ( –0.41)

= +0.81 V > 0 [1]


O2

(g) + 2H2O (l ) + 4Cr 2+

(aq) 4Cr 3+

(aq) + 4OH

(aq) [1] and Cr 3+ (aq) + 3OH (aq) Cr(OH)3 (s) [1]

Blue solution decolourises and a greyish –green ppt is formed in a green solution. [1]

3(a) O2 (g) + 4H+ (aq) + 4e 2H2O (l ) [1]

(b) CH4 (g) + 2H2O (l ) CO2 (g) + 8H+ (aq) + 8e [1]

(c) CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l ) [1]

(d) So long as the two reactants i.e. O2(g) and CH4(g) are not in direct contact [1], there isno need to have two separate fuel cells. (These two reactants are in fact, “separated” by acommon electrolyte i.e. aqueous dilute H2SO4.)

(e)

(f) CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) [1]

(g) They are the same except for the state of water. [1]

(h) More efficient conversion of chemical energy into electrical energy. [1] Also the products formed are much less polluting.

4(a) MnO4 –(aq) + 8H+(aq) + 5e ⇌ Mn2+(aq) + 4H2O (l ) E = +1.52 V

Cl 2(g) + 2e ⇌ 2Cl (aq) E = +1.36 V

Cr 2O7(aq) + 14 H+(aq) + 6e ⇌ 2Cr 3+(aq) + 7H2O(l ) E = +1.33 V

From the E values,it can be seen that, under standard conditions, MnO4

can oxidise Cl to Cl i.e. Ecell = +1.52 – (+1.36)

= +0.16 V > 0 [1]

In the titration determination of iron(II) ions (e.g. using aqueous FeCl 2), the amount ofFe2+ present is determined by the amount of oxidant it reacts with.

Pt

V

Pt

O2 (g)

(cathode)

dilutesulphuricacid

CH4 (g)

(anode)

e – Direction of e – flow [1]

Labelling of electrodes [1] Labelling of electrolyte and gases [1]


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If MnO4 were used, some MnO4

will be consumed by Cl instead of reacting solely withFe2+. [1] This would result in a larger amt of MnO4

– used than required i.e. the resultswould be inaccurate.

On the other hand, under standard conditions, Cr 2O72 cannot oxidise Cl to Cl 2

i.e. Ecell = +1.33 – (+1.36)= -0.03 V < 0 [1]

Hence all Cr 2O72

will be used to react with Fe2+

and thus K2Cr 2O7 (aq) can be used forthe titration determination of iron(II) ions in the presence of chloride ions

4(b) Al 3+ + 3e – ⇌ Al E = –1.66 V

Pb2+ + 2e – ⇌ Pb E = –0.13 V

Aluminium has a very low reduction potential and so its ion is not easily reduced to themetal. [1]Its ore (e.g. Al 2O3) cannot be reduced by C and requires electrolysis. [1]

(Note: the Al 2O3 ore is not soluble in water and even if it were dissolved in acid to make up anaqueous solution, the latter would generate H2 (from water) at the cathode. Hence molten ore isneeded.)

Lead has a less negative reduction potential and so it is relatively easier for its ion tobe reduced to the metal. [1] Hence its oxide can be reduced by C.

(Note: carbon is a much cheaper raw material compared to the expensive electrical power neededfor electrolysis.)

(c) At the cathode: 2H2O(l ) + 2e – H2(g) + 2OH –(aq) [1] H2 (E = –0.83 V) is preferentially discharged [1] since Na

+ (E= –2.71 V) is lessreadily reduced.

At the anode,

gas B contains 80% C by mass high C content; other element present could be H.

Assuming B is CxHy, then for 1.0 g of B,

mole ratio of C:H =0.1

20.0:

0.12

80.0 = 1:3 i.e. empirical formula = CH3

B is likely C2H6. [1]

The oxidation of CH3CO2 –(aq) could likely produce CO2 as the other gas A. [1]

Hence at anode, the balanced half –equation is:2CH3CO2

–(aq) CH3CH3(g) + 2CO2(g) + 2e – [1]

From the half –equation, it can be seen that the amt of CO2 is twice that of CH3CH3. At the same temperature and pressure, the volume ratio of CO2: CH3CH3 is 2:1. [1]


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5(a)(i) amt of Ag deposited = 0.100 / 108

= 9.26 x 10 –4 mol [1]

(ii) Ag+(aq) + e – Ag

amt of e – = amt of Ag deposited

= 9.26 x 10 –4 mol [1]

quantity of charge passed = 96 500 x 9.26 x 10 –4 = 89.35 C

current passed = 89.35 / (30 x 60)

= 0.0496 A

0.05 A [1]

(iii) Cr 3+(aq) + 3e Cr (s)

amt of Cr deposited = 1/3 x amt of e – passed

= 3.086 x 10 –4 mol [1]

mass of Cr deposited = 3.086 x 10

–4

x 52.0= 0.0160 g [1]

(b) volume of Al 2O3 to be deposited = 500 x 1 x 10 –3

= 0.5 cm3

mass of Al 2O3 = 4.0 x 0.5

= 2.0 g [1]

amt of Al 2O3 = 2.0 / (27.0 x 2 + 16.0 x 3)

= 0.0196 mol

At the anode: 2H2O(l ) O2(g) + 4H+(aq) + 4e

then 4Al (s) + 3O2(g) 2Al 2O3 (s) [1]

amt of e – = 4 x amt of O2 produced

= 4 x (2

3 x amt of Al 2O3)

= 0.1176 mol

quantity of charge needed = 96 500 x 0.1176= 11 348 C [1]

(c) Ni2+ (aq) + 2e Ni (s)

2Ag+ (aq) + 2e 2Ag (s)

For the same current passed,

amt of Ag: amt of Ni deposited = 2 : 1 [1]

hence mass of Ag: mass of Ni deposited = 2 x 108 : 1 x 58.7

= 3.68 : 1 [1]


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5(d) cathode: 2Cu2+ (aq) + 4e 2Cu (s)

anode: 2H2O(l ) O2(g) + 4H+(aq) + 4e

amt of O2 produced = ½ x amt of Cu deposited [1]

= ½ x (0.635 / 63.5)

= 0.005 mol [1]

vol. of O2 produced at r.t.p. = 0.005 x 24.0

= 0.12 dm3 [1]

Addtional Questions (Optional)

A1(i) Ecell = +0.77(+0.34) = +0.43 V [1]

(ii) Fe3+ (aq) + e Fe2+ (aq) [1]Cu (s) Cu2+ (aq) + 2e [1]

2Fe3+

(aq) + Cu (s) 2Fe2+

(aq) + Cu2+

(aq) [1]

(iii) Ecell of the reaction is +0.43V >>0. Hence even though the conditions in a test –tube maynot be under standard conditions, Ecell is still > 0 such that reaction can proceed in a test –tube. [1]

(iv) (Any suitable half –cell with E > +0.77 V) e.g.Cl 2 / Cl

– half –cell: E = + 1.36V [1]Electrode: Pt, with chlorine gas and sodium chloride solution. [1]

A2(i) O2 (g) + 2H2O (l ) + 4e

⇌ 4OH

(aq) E = +0.40 V Zn2+ (aq) + 2e ⇌ Zn (s) E = –0.76 V O2(g) + 2H2O (l ) + 2Zn (s) 2Zn(OH)2 (s) [1]

Ecell = +0.40(+0.76) = +1.16 V [1]

(ii)

(iii) aqueous KCl . [1]

(Note: choice of electrolyte requires careful consideration i.e. it must not oxidise Zn or C.e.g. dilute acids cannot be used since H+ reacts with Zn.

solutions containing ions of metals less reactive than Zn (e.g. CuSO4) cannot beused as it undergoes displacement reaction with Zn.)

Zn

electrolyte (aq)

C

V

e

O2 (g)

( –) (+)


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Self –Check

1(i) The standard electrode potential of a half –cell is the electromotive force, measured at 298Kand 1 atm, between the half –cell and the standard hydrogen electrode, in which the reactingspecies in solution are at molar concentrations.

(ii) Ecell is the potential difference between two half –cells under standard conditions. It gives ameasure of the e.m.f. (electromotive force) which “pumps” the electrons around the circuit.

2(i)

(ii)

3(i)E of Co2+ / Co = - 0.28V (oxid) Co is negative electrode

E of Cl 2 / Cl ¯ = + 1.36V (red) Pt is positive electrode

(ii) Cl 2 (g) + Co(s) 2Cl ¯(aq) + Co2+(aq)

(iii) Ecell = 1.36 – (-0.28) = +1.64V

V

T = 298 K

[H+(aq)] = 1 mol dm –3

H+

H2 (1 atm)

Pt

[Fe2+(aq)] = [Fe3+(aq)]= 1 mol dm –3

Pt

salt bridge

Fe3+

Fe2+

[H+(aq)] = 1 mol dm –

H+

H2 (1 atm)

Pt

Cu2+

Cu

[Cu2+(aq)] = 1 mol dm –3

V

T = 298 K

salt bridge

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Electrochemistry [Ans]