Standard electrode potentials and thermodynamic feasibility

Question 1

using standard electrode potentials, E⦵ values, is one way of

Answer 1

• measuring how easily a species loses electrons
• this method provides information on how far to the left an equilibrium is relative to the equilibrium in the standard hydrogen electrode

Question 2

use the following two equilibria and their E⦵ values to how relatively their equilibria are far left to the standard hydrogen electrode:

Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 V

Cu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 2

The position of equilibrium of the Zn2+(aq) | Zn(s) reaction lies further to the left than that of the Cu2+(aq) | Cu(s) reaction as the Zn2+(aq) | Zn(s) E⦵ value if more negative than the Cu2+(aq) | Cu(s) E⦵ value

Question 3

if the following equilibria are linked by combining the two half-cells to make an electrochemical cell, the electrons will flow from the …………..electrode to the …………….electrode
Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 V

Cu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 3

e- flow from the zinc electrode to the copper electrode because zinc is the negative electrode, having the more negative E⦵ value for the half-cell

electrons flow from the half-cell with the more negative E⦵ value to the half-cell with the less negative E⦵ value

Question 4

explain whether the following reaction thermodynamically feasible:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

use the following info to help you:
Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 V
Cu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 4

• The E⦵ value for the Zn equilibrium is more negative than the Cu equilibrium (-0.76V to +0.34V)
• so the position of the Zn equilibrium will shift to the left, releasing electrons while the position of the Cu will shift to the right, accepting electrons
• this means the reaction between Zn(s) and Cu2+(aq) is thermodynamically feasible

Question 5

the method for working out, form half ionic equations, how to construct a full equation is:

Answer 5

list both ionic equation in their order of reducing power (most negative on top, and most positive value on the bottom)
the top right species will react with the bottom left species as follows:

Zn2+(aq) + 2e- ⇌ {Zn(s)} E⦵ = -0.76 V
{Cu2+(aq)} + 2e- ⇌ Cu(s) E⦵ = +0.34 V

the species inside the {* *} will react together in what is called the anti-clockwise rule

another way of looking at this is that the MOST negative E⦵ value has its equilibrium moving to the left (so you need to flip that equation over):
Zn(s) ⇌ 2e- + Zn2+(aq)
giving away the full equation as:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Question 6

use the data below to deduce whether zinc will react with dilute sulfuric acid:
Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ = -0.76 V
2H+(aq) + 2e- ⇌ H2(g) E⦵ = 0.00 V

Answer 6

• with the half equations arranged in order of reducing power (most negative on top, most positive on bottom), you can see Zn(s) reacts with H+
• therefore the equation for the reaction is:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
• calculating the E⦵cell value for the reaction gives:
E⦵cell = E⦵red - E⦵oxi (E⦵RHS - E⦵LHS)

E⦵cell = 0.00 - -0.76 = +0.76V
• Zinc releases e- to the hydrogen ions
• The Reaction is thermodynamically feasible

Question 7

use the data below to deduce whether copper will react with dilute sulfuric acid:
2H+(aq) + 2e- ⇌ H2(g) E⦵ = 0.00 V
Cu2+(aq) + 2e- ⇌ Cu(s) E⦵ = +0.34 V

Answer 7

• With the half equation arranged in order of reducing power, we can see that Cu(s) will NOT react with H+ ions as the copper equilibrium is more positive, meaning:
Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
is not feasible because:
E⦵cell = 0.00 - 0.34 = -0.34 V
• E⦵cell is negative, so the reaction is not thermodynamically feasible

instead, the reaction:
Cu2+ + H2(g) → Cu(s) + 2H+(aq) is feasible, but has a very large activation energy, so the reactants are said to be kinetically stable

Question 8

one way of making chlorine in the lab is to

Answer 8

react manganese(IV) oxide with HCl acid

Question 9

the equation for the reaction between manganese(IV) oxide and hydrochloric acid is:

Answer 9

MnO2(s) + 4HCl(aq) → Mn2+(aq) + 2Cl-(aq) + 2H2O(l) + Cl2(g)

Question 10

the two half equations that lead to the following reaction are:
MnO2(s) + 4HCl(aq) → Mn2+(aq) + 2Cl-(aq) + 2H2O(l) + Cl2(g)

Cl2(g) + 2e- ⇌ 2Cl-(aq)
E⦵ = +1.23 V
MO2(s) + 4H+(aq) + 2e- ⇌ Mn2+(aq) + 2H2O(l)
E⦵ = +1.36 V

Answer 10

-calculating the E⦵cell value for the reaction gives:
E⦵cell = E⦵red -E⦵oxi (E⦵RHS - E⦵LHS)

E⦵cell = 1.23 - 1.36 = -0.13 V
-the chlorine cannot release e- to the MnO2
-this means the reaction is not thermodynamically feasible under standard conditions, BUT could happen if the concentration of the HCl (H+ ions and Cl- ions) is increased to more than 1 mol dm-3
-this shifts the position of equilibrium below to the right:
MO2(s) + 4H+(aq) + 2e- ⇌ Mn2+(aq) + 2H2O(l)
-and the position of equilibrium below to the left:
Cl2(g) + 2e- ⇌ 2Cl-(aq)

• as a result, the E value for the Mn equilibrium becomes more positive as the equilibrium is forced towards the right, meaning the redox system becomes a better electron acceptor
• the Cl equilibrium becomes a better e- releaser meaning its E value becomes less positive

-the net effect is that the Cl equilibrium’s E value is now less positive (more negative) than the Mn equilibrium, so the reaction is thermodynamically feasible in these non-standard conditions

Question 11

a disproportionation reaction is

Answer 11

one in which an element in a species is simultaneously oxidised and reduced in the same reaction

Question 12

the equation for the disproportionation of Cu+ ions is:

Answer 12

2Cu+(aq) → Cu2+(aq) + Cu(s)

Question 13

explain how the half cells below result in a disproportionation reaction:

Cu2+(aq) + e- ⇌ Cu+(aq)
E⦵ = +0.15 V
Cu+(aq) + e- ⇌ Cu(s)
E⦵ = +0.52 ⦵

Answer 13

Cu2+(aq) + e- ⇌ Cu+(aq)
E⦵ = +0.15 V
Cu+(aq) + e- ⇌ Cu(s)
E⦵ = +0.52 ⦵
-the E⦵ value for the Cu2+(aq) equilibrium is more negative than for the Cu(s) equilibrium, meaning the Cu2+(aq) equilibrium position lies more to the left and the position of equilibrium for Cu(s) lies more to the right
-this means the Cu+(aq) ions will both release and accept electrons to one another to form Cu2+(aq) and Cu(s)

Question 14

the relationship between ΔG⦵ and the emf of the cell, E⦵cell, is

Answer 14

ΔG⦵ = -nFE⦵cell

where n is the number of moles of electrons involved in the cell reaction and F is the Faraday constant

Question 15

ΔG⦵ = (in terms of entropy and enthalpy)

Answer 15

ΔG⦵ = ΔH - TΔSsystem

but Δsurroundings = -ΔH / T
therefore:
ΔH = -TΔSsurroundings

so ΔG = -TΔSsussoundings - TΔSsystem
= -T (ΔSsurroundings + ΔSsystem)
= -TΔStotal

Question 16

-TΔStotal =

Answer 16

as ΔG = -TΔStotal
and ΔG = -nFE⦵cell

then
-TΔStotal = -nFE⦵cell
so
TΔStotal = nFE⦵cell

Question 17

as F and n are constants in TΔStotal = nFE⦵cell , then TΔStotal is:

Answer 17

then ΔS⦵total ∝ E⦵cell

Question 18

if E⦵cell is positive, then

Answer 18

then ΔS⦵total will be positive and so the reaction for the cell diagram as written from left to right will be thermodynamically feasible

Question 19

use ΔS⦵total ∝ E⦵cell to deduce whether the following reaction is feasible:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

E⦵cell = +1.10 V

Answer 19

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) 
E⦵cell = +1.10 V
here E⦵cell is positive, so using:
 ΔS⦵total ∝ E⦵cell 
we can see how ΔS⦵total will also be positive, meaning the reaction is thermodynamically stable, so the reaction below will go ahead in standard conditions

the half cell reactions are:
Zn(s) → Zn2+(aq) + e-
Cu2+ + 2e- → Cu(s)
so the overall equation for the reaction is:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

if E⦵cell was negative for the half-cell diagram, then the reaction would occur from right to left, so that equations would be flipped over to make the equation for a thermodynamically feasible reaction

Question 20

use ΔS⦵total ∝ E⦵cell to deduce whether the following reaction is feasible:
Pt(s) | 1/2 H2(g) | H+(aq) || Zn2+(aq) | Zn(s)
E⦵cell = -0.76 V

Answer 20

Pt(s) | 1/2 H2(g) | H+(aq) || Zn2+(aq) | Zn(s)
E⦵cell = -0.76 V
here E⦵cell is negative, so the half-cell reactions must be written from right to left for the reaction to occur:

Zn(s) → Zn2+(aq) + e-
H+(aq) + e- → 1/2 H2(g)
so the overall reaction is:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Question 21

ΔG⦵ is related to the equilibrium constant, K, through the expression:

Answer 21

ΔG⦵ = -RTlnk

Question 22

use ΔG⦵ = -RTlnk to show the relationship between lnK and E⦵cell

Answer 22

ΔG = -nFE⦵cell
therefore
RTlnK = nFE⦵cell
or
lnK = nFE⦵cell / RT

as F, n and R are constants, then at a given temperature, T, lnK is proportional to E⦵cell
lnK ∝ E⦵cell

E⦵cell can therefore be used to calculate the thermodynamic equilibrium constant, K, for a cell reaction

Question 23

lnK =

Answer 23

lnK = nFE⦵cell / RT

as nFE⦵cell = TΔStotal , then

lnK = (T)ΔStotal / R(T)

lnK = ΔStotal / R