Tutorial 20: Electrochemistry – Suggested Solutions 1(a)(1)
Cr 2O72 – (aq) + 14H+ (aq) + 6e
(i)
Cu2+(aq) + 2e-
⇌
2Cr 3+ (aq) + 7H2O (l )
E = +1.33 V
⇌
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 )
Anode: Oxid:
[1]
Copper electrode in Cu2+ (aq) / Cu (s) half –cell
Cu (s)
Cu2+ (aq) + 2e
[1]
(ii)
Cr 2O72 – (aq) + 14H+(aq) + 3Cu(s) 2Cr 3+ (aq) + 7H2O(l ) + 3Cu2+(aq)
[1]
(iii)
Ecell = +1.33 – (+0.34) = +0.99 V
[1]
(iv)
e –
Correct drawing & labeling Correct conditions Correct e – direction
V
T = 298 K
[1] [1] [1]
Cu
salt bridge Pt
[Cr 2O7 – (aq)] = [Cr 3+(aq)] = [H+(aq)] = 1 mol dm –3
+
[Cu (aq)] = 1 mol dm –3
1(a)(2) (i)
H2O2 (aq) + 2H+ (aq) + 2e O2 (g) + 2H+ (aq) + 2e
2H2O (l )
E = +1.77 V
⇌
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: Oxid:
[1]
Pt electrode in O2 (g) / H2O2 (aq) half –cell
H2O2 (aq) O2 (g) + 2H+ (aq) + 2e
(ii)
2H2O2 (aq) 2H2O (l ) + O2 (g)
[1]
(iii)
Ecell = +1.77 – (+0.68) = +1.09 V
[1]
(iv)
e –
Correct drawing & labeling Correct conditions Correct e – direction
V
T = 298 K O2 (1 atm) [H2O2(aq)] = [H+(aq)] = 1 mol dm –3
salt bridge Pt
Pt
1
[H2O2(aq)] = [H+(aq)] = 1 mol dm –3
[1] [1] [1]
1(b)(i)
CO32 - precipitates out Cu 2+ so that [Cu2+(aq)] decreases.
[1]
This shifts the position of equilibrium in Cu 2+(aq) + 2e Cu(s) to the left so that EL decreases. Hence E cell = ER – EL increases (i.e. Ecell > E cell). [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 E R or increase E L. (1) ER can be decreased by using H 2O2 or H + of a lower concentration in the H 2O2/H2O half –cell. [1] + (2) EL can be increased by using O 2 of a higher pressure, H of a higher concentration or H2O2 of a lower concentration in the O 2 / H2O2 half –cell. [1]
(ii)
Hydrogen peroxide can spontaneously undergo oxidation and reduction under standard conditions: 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 E R > EL.)
2(a)
I2 (aq) + 2e 2I (aq) Fe2+ (aq) + 2e Fe (s)
E = +0.54 V E = –0.44 V
⇌
⇌
Ecell = –0.44 – (+0.54) = –0.98 V < 0 reaction is not likely to occur. (b)
H2O2 (aq) + 2H+ (aq) + 2e
I2 (aq) + 2e
[1]
2H2O (l )
E = +1.77 V E = +0.54 V
⇌
2I (aq)
⇌
Ecell = +1.77 – (+0.54) = +1.23 V > 0 [1] reaction is energetically feasible and spontaneous under standard conditions. H2O2 (aq) + 2H+ (aq) + 2I (aq) 2H2O (l ) + I2 (aq) A brown solution is formed.
(c)
Cr 2O72 – (aq) + 14H + (aq) + 6e
SO42 – (aq) + 4H+ (aq) + 2e
-
⇌
2Cr 3+ (aq) + 7H2O (l ) SO2 (aq) + 2H2O (l )
⇌
[1] [1]
E = +1.33 V E = +0.17 V
Ecell = +1.33 – (+0.17) = +1.16 V > 0 [1] reaction is energetically feasible and spontaneous under standard conditions. Cr 2O72 –(aq) + 2H+ (aq) + 3SO 2 (aq) 2Cr 3+ (aq) + H2O (l) + 3SO42 – (aq) Orange solution turns green. 2
[1] [1]
2(d)
O2 (g) + 2H2O (l ) + 4e
Cr 3+ (aq) + e
⇌
4OH (aq) Cr 2+ (aq)
E = +0.40 V E = –0.41 V
⇌
Ecell = +0.40 – ( –0.41) = +0.81 V > 0 [1] reaction is energetically feasible and spontaneous under standard conditions. O2 (g) + 2H2O (l ) + 4Cr 2+ (aq) 4Cr 3+(aq) + 4OH (aq) and Cr 3+ (aq) + 3OH (aq) Cr(OH)3 (s) Blue solution decolourises and a greyish –green ppt is formed in a green solution.
[1] [1] [1]
3(a)
O2 (g) + 4H+ (aq) + 4e
(b)
CH4 (g) + 2H2O (l ) CO2 (g) + 8H+ (aq) + 8e
(c)
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l )
(d)
So long as the two reactants i.e. O 2(g) and CH4(g) are not in direct contact [1], there is
2H2O (l )
[1]
[1] [1]
no need to have two separate fuel cells. (These two reactants are in fact, “separated” by a
common electrolyte i.e. aqueous dilute H 2SO4.) (e)
Direction of e – flow Labelling of electrodes Labelling of electrolyte and gases
–
e
V CH4 (g)
O2 (g)
(anode)
(cathode)
Pt
Pt
[1] [1] [1]
dilute sulphuric acid
(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. Also the products formed are much less polluting.
[1]
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, MnO 4 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 FeC l 2), the amount of Fe2+ present is determined by the amount of oxidant it reacts with. 3
If MnO4 were used, some MnO 4 will be consumed by C l instead of reacting solely with Fe2+. [1] This would result in a larger amt of MnO 4 – used than required i.e. the results would be inaccurate.
On the other hand, under standard conditions, Cr 2O72 cannot oxidise C l 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 Fe 2+ and thus K2Cr 2O7 (aq) can be used for the titration determination of iron(II) ions in the presence of chloride ions
4(b)
Al 3+ + 3e – Pb2+ + 2e –
⇌ ⇌
Al Pb
E = –1.66 V E = –0.13 V
Aluminium has a very low reduction potential and so its ion is not easily reduced to the metal. [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 an aqueous solution, the latter would generate H2 (from water) at the cathode. Hence molten ore is needed.)
Lead has a less negative reduction potential and so it is relatively easier for its ion to be 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 needed for 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 less readily 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
=
0.80 12.0
:
0.20 1.0
= 1:3 i.e. empirical formula = CH3
B is likely C2H6.
[1]
The oxidation of CH 3CO2 –(aq) could likely produce CO2 as the other gas A. Hence at anode, the balanced half –equation is: 2CH3CO2 –(aq) CH3CH3(g) + 2CO2(g) + 2e –
[1] [1]
From the half –equation, it can be seen that the amt of CO 2 is twice that of CH3CH3. At the same temperature and pressure, the volume ratio of CO 2: CH3CH3 is 2:1. [1]
4
5(a)(i) amt of Ag deposited
(ii)
Ag+(aq) + e – Ag amt of e – quantity of charge passed current passed
= 0.100 / 108 = 9.26 x 10 –4 mol
= amt of Ag deposited = 9.26 x 10 –4 mol = 96 500 x 9.26 x 10 –4 = 89.35 C = 89.35 / (30 x 60) = 0.0496 A
(iii)
Cr 3+(aq) + 3e Cr (s) amt of Cr deposited
[1]
[1]
mass of Cr deposited
(b)
0.05 A
[1]
= 1/3 x amt of e – passed = 3.086 x 10 –4 mol = 3.086 x 10 –4 x 52.0 = 0.0160 g
volume of Al 2O3 to be deposited mass of Al 2O3 amt of Al 2O3
= 500 x 1 x 10 –3 = 0.5 cm3 = 4.0 x 0.5 = 2.0 g = 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) amt of e –
[1] [1]
[1]
[1]
= 4 x amt of O2 produced =4x(
3 2
x amt of A l 2O3)
= 0.1176 mol quantity of charge needed
(c)
= 96 500 x 0.1176 = 11 348 C
Ni2+ (aq) + 2e Ni (s) 2Ag+ (aq) + 2e 2Ag (s) For the same current passed, amt of Ag: amt of Ni deposited hence mass of Ag: mass of Ni deposited
[1]
=2:1 = 2 x 108 : 1 x 58.7 = 3.68 : 1
5
[1] [1]
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 = ½ x (0.635 / 63.5) = 0.005 mol
vol. of O2 produced at r.t.p.
= 0.005 x 24.0 = 0.12 dm3
[1] [1]
[1]
Addtional Questions (Optional)
A1(i)
Ecell
= +0.77(+0.34)
= +0.43 V
[1]
(ii)
Fe3+ (aq) + e Fe2+ (aq) Cu (s) Cu2+ (aq) + 2e 2Fe2+ (aq) + Cu2+ (aq)
[1] [1] [1]
2Fe3+ (aq) + Cu (s) (iii)
Ecell of the reaction is +0.43V >>0. Hence even though the conditions in a test –tube may not 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 Electrode: Pt, with chlorine gas and sodium chloride solution.
A2(i)
O2 (g) + 2H2O (l ) + 4e 4OH (aq) Zn2+ (aq) + 2e Zn (s)
⇌
[1] [1]
E = +0.40 V E = –0.76 V
⇌
O2(g) + 2H2O (l ) + 2Zn (s)
2Zn(OH)2 (s)
[1]
Ecell
= +1.16 V
[1]
= +0.40(+0.76)
(ii)
V e
O2 (g)
Zn
( –)
(+)
C
electrolyte (aq)
(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. CuSO 4) cannot be used as it undergoes displacement reaction with Zn.)
6
–Check Self 1(i) The standard electrode potential of a half –cell is the electromotive force, measured at 298K and 1 atm, between the half –cell and the standard hydrogen electrode, in which the reacting species in solution are at molar concentrations. (ii) Ecell is the potential difference between two half –cells under standard conditions. It gives a measure of the e.m.f. (electromotive force) which “pumps” the electrons around the circuit. 2(i)
V
T = 298 K H2 (1 atm)
Cu
salt bridge Pt Cu2+ H+
[H+(aq)] = 1 mol dm –
[Cu2+(aq)] = 1 mol dm –3
(ii) V
T = 298 K H2 (1 atm)
salt bridge Pt
Pt Fe2+ Fe3+
H+
[H+(aq)] = 1 mol dm –3
[Fe2+(aq)] = [Fe3+(aq)] = 1 mol dm –3
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
7
