Questions on electrode potentials and the standard hydrogen electrode

Question 1

Explain why aluminium foil placed in copper(II) sulfate solution shows almost no change for several minutes, yet becomes coated with copper, and the mixture warms, soon after sodium chloride is dissolved in the solution. E°(Al³⁺/Al) = −1.66 V; E°(Cu²⁺/Cu) = +0.34 V. Your answer should predict whether the reaction is feasible, give its equation, and explain why a feasible reaction may not be seen.

Accepted Answer

Indicative content

copper(II) ions are reduced and aluminium is oxidised, so the copper half-cell goes on the right and E°cell = 0.34 − (−1.66) = +2.00 V
E°cell is positive, so the reduction of Cu²⁺ by aluminium is feasible under standard conditions and copper is predicted to form
the equation balances six electrons: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), with neither E° value multiplied
E° values predict whether a reaction can happen, not how fast, because they say nothing about the activation energy of the reaction
the aluminium is covered by a thin, unreactive layer of aluminium oxide that keeps the copper(II) ions off the metal, so at first the rate is negligible
chloride ions break down the oxide layer and expose the metal, so the feasible reaction now goes at a visible rate and the energy it releases warms the mixture

Levels

Level 3 (5–6 marks) E°cell is calculated correctly, the reaction is predicted feasible, the balanced six-electron equation is given, the limit of E° values to feasibility rather than rate is stated, and the oxide layer and its removal by chloride are used to explain both the delay and the later reaction
Level 2 (3–4 marks) E°cell is calculated, the reaction predicted feasible and the equation is right; the delay is explained only in part, such as the oxide layer with no role for chloride, or E° values are not separated from rate
Level 1 (1–2 marks) the reaction is called feasible or the oxide layer mentioned; the E° values are typically taken to mean the reaction must start at once, and the equation does not balance the electrons

Question 2

Describe how you would measure the EMF of a cell made from a zinc half-cell and a copper half-cell in a school laboratory.

Accepted Answer

Indicative content

clean the zinc and copper strips with emery paper to remove the oxide layer, then rinse and dry them
stand the zinc in 1.00 mol dm⁻³ zinc sulfate and the copper in 1.00 mol dm⁻³ copper(II) sulfate, in separate beakers
join the solutions with a strip of filter paper soaked in saturated potassium nitrate solution as a salt bridge
connect the two metals to a high-resistance voltmeter, positive terminal to the copper, so that almost no current flows
read the voltmeter once the reading is steady, and record the temperature, which should be 298 K
repeat with fresh solutions and a fresh salt bridge, and take the mean of concordant readings

Levels

Level 3 (5–6 marks) all the elements of a standard measurement are present: cleaned electrodes, 1.00 mol dm⁻³ solutions of each metal's own ions, a salt bridge and what it is made of, a high-resistance voltmeter with the polarity given, a steady reading at 298 K and repeats
Level 2 (3–4 marks) the two half-cells, the salt bridge and the voltmeter are described correctly; the 1.00 mol dm⁻³ concentrations, the temperature, the high-resistance voltmeter or the cleaned electrodes are not mentioned
Level 1 (1–2 marks) zinc and copper are dipped in solutions joined to a voltmeter, often in the same beaker or with a wire in place of the salt bridge

Question 3

Calculate the electrode potential of each metal half-cell from a student's readings, given in the table, for three cells measured by the method of required practical 8, each with a copper half-cell as the positive electrode. E°(Cu²⁺/Cu) = +0.34 V. Use your values to predict the EMF of a cell made from the zinc and nickel half-cells, stating which electrode is positive.

Accepted Answer

M1 copper is the right-hand electrode, so E(metal) = 0.34 − reading A1 E(Ni²⁺/Ni) = −0.23 V, E(Pb²⁺/Pb) = −0.12 V and E(Zn²⁺/Zn) = −0.75 V M1 for the zinc–nickel cell, EMF = E(Ni) − E(Zn) = (−0.23) − (−0.75); allow the difference of the two readings, 1.09 − 0.57 B1 nickel has the less negative potential, so the nickel electrode is positive and goes on the right A1 predicted EMF = +0.52 V; allow error carried forward from the electrode potentials

Question 4

Describe the standard hydrogen electrode, giving the conditions under which it is used, and explain why its electrode is made of platinum.

Accepted Answer

B1 the solution contains H⁺ ions at a concentration of 1.00 mol dm⁻³, for example 1.00 mol dm⁻³ hydrochloric acid (allow 1 mol dm⁻³) B1 hydrogen gas is bubbled in at 100 kPa and the temperature is 298 K B1 the gas passes over a platinum electrode, coated with platinum black, that dips into the acid B1 platinum conducts and is inert, so it carries electrons to and from the H⁺/H₂ equilibrium without reacting, since hydrogen cannot itself be an electrode

Question 5

Deduce which two of the half-cells in the table, connected together, give the cell with the largest EMF, and calculate that EMF. Then write the conventional representation of the cell with the smallest EMF. The table gives four standard electrode potentials.

Accepted Answer

B1 the largest EMF comes from the two values furthest apart: magnesium and silver, with silver on the right M1 E°cell = E°right − E°left = (+0.80) − (−2.37) A1 E°cell = +3.17 V B1 smallest EMF is copper with silver, +0.46 V: Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s); reject zinc with copper, +1.10 V; ignore the EMF if not given

Question 6

Explain why the EMF of a cell is measured with a high-resistance voltmeter, and why a metal wire cannot be used in place of the salt bridge.

Accepted Answer

B1 a high-resistance voltmeter allows almost no current to flow through the cell B1 no voltage is lost driving a current and the concentrations do not change, so the reading is the EMF; reject 'more accurate' alone B1 each end of a wire dipping into a solution would form a new metal–solution half-cell with its own electrode potential B1 those extra potentials would add to the reading, so the measured value would not be the EMF of the cell

Question 7

Calculate E°(Ni²⁺/Ni) from the standard cell in the diagram, given that E°(Cu²⁺/Cu) = +0.34 V. Then calculate the standard EMF of a cell made from a nickel half-cell and a silver half-cell, E°(Ag⁺/Ag) = +0.80 V.

Accepted Answer

M1 E°cell = E°right − E°left, so +0.59 = (+0.34) − E°(Ni²⁺/Ni) A1 E°(Ni²⁺/Ni) = 0.34 − 0.59 = −0.25 V M1 silver is more positive and goes on the right: E°cell = (+0.80) − (−0.25); allow error carried forward from the nickel value A1 E°cell = +1.05 V

Question 8

Give the meaning of the term standard electrode potential.

Accepted Answer

B1 the EMF, or potential difference, of a cell measured when no current flows (allow 'voltage') B1 in which the half-cell is connected to a standard hydrogen electrode; reject an answer with no reference electrode B1 with both half-cells under standard conditions: 298 K, 100 kPa and 1.00 mol dm⁻³ of each ion

Question 9

Explain the function of the salt bridge in an electrochemical cell, and explain why potassium nitrate is a suitable salt to use in it.

Accepted Answer

B1 the salt bridge allows ions to move between the two half-cells, completing the circuit; reject 'carries electrons' B1 the movement of ions keeps each solution electrically neutral without the two solutions mixing B1 potassium ions and nitrate ions do not react with the ions in either half-cell and form no precipitate

Question 10

Explain how a potential difference arises between a strip of zinc and a solution of zinc ions in which it stands.

Accepted Answer

B1 an equilibrium is set up at the surface of the metal, Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) B1 zinc atoms leave the metal as Zn²⁺ ions and leave their electrons behind on the strip B1 the metal becomes negatively charged relative to the solution, giving a potential difference between them

Question 11

Calculate the standard EMF of a cell made from a standard zinc half-cell, E° = −0.76 V, and a standard silver half-cell, E° = +0.80 V, and state which electrode is positive.

Accepted Answer

M1 silver has the more positive E°, so the silver half-cell is the right-hand, positive electrode M1 E°cell = E°right − E°left = (+0.80) − (−0.76) A1 E°cell = +1.56 V, with the silver electrode positive

Question 12

Calculate the standard EMF of the cell Mg(s) | Mg²⁺(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s), and write the equation for the reaction when it delivers a current. E°(Mg²⁺/Mg) = −2.37 V; E°(Fe³⁺/Fe²⁺) = +0.77 V.

Accepted Answer

M1 E°cell = E°right − E°left = (+0.77) − (−2.37) B1 Mg(s) + 2Fe³⁺(aq) → Mg²⁺(aq) + 2Fe²⁺(aq), with the iron half-equation doubled and its E° not doubled A1 E°cell = +3.14 V

Question 13

Complete the conventional representation of the cell made from a standard Fe³⁺/Fe²⁺ half-cell, E° = +0.77 V, and a standard silver half-cell, E° = +0.80 V, and give its EMF.

Accepted Answer

B1 the silver half-cell is on the right, written || Ag⁺(aq) | Ag(s), because its E° is more positive B1 the iron half-cell is on the left, written Pt(s) | Fe²⁺(aq), Fe³⁺(aq) ||, with platinum at the outer end, a comma between the ions and Fe³⁺ next to the double line B1 EMF = 0.80 − 0.77 = +0.03 V, with the silver electrode positive

Question 14

Suggest how the standard electrode potential of the Cl₂/Cl⁻ half-cell could be measured, describing the chlorine half-cell.

Accepted Answer

B1 chlorine gas at 100 kPa bubbled over a platinum electrode B1 in a solution of chloride ions at 1.00 mol dm⁻³, such as potassium chloride solution, at 298 K B1 connected by a salt bridge and a high-resistance voltmeter to a standard hydrogen electrode; the reading is E°, with the chlorine electrode positive

Question 15

Compare a standard zinc half-cell with a standard Fe³⁺/Fe²⁺ half-cell, in terms of the species present and the electrode used.

Accepted Answer

B1 the zinc half-cell is a metal in a solution of its own ions, and the zinc metal is the electrode B1 the iron half-cell holds two ions, Fe³⁺ and Fe²⁺, in the same solution, each at 1.00 mol dm⁻³ B1 neither iron species is a solid, so an inert platinum electrode is needed; in both, a redox equilibrium is set up at the electrode surface

Question 16

Evaluate the statement that the salt bridge carries electrons from one half-cell to the other.

Accepted Answer

B1 the statement is wrong: electrons flow only through the external wire and the electrodes B1 ions move through the salt bridge, anions towards the half-cell where oxidation occurs and cations towards the one where reduction occurs B1 that movement of ions keeps each solution electrically neutral, so the current can continue

Question 17

Justify the choice of the zinc electrode as the negative pole of a cell made from a standard zinc half-cell, E° = −0.76 V, and a standard copper half-cell, E° = +0.34 V.

Accepted Answer

B1 zinc has the more negative E°, so the equilibrium Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) lies further to the left than the copper one and zinc releases electrons more readily B1 zinc is oxidised, Zn(s) → Zn²⁺(aq) + 2e⁻, and the electrons it leaves on the metal make it the negative electrode; electrons flow from zinc to copper through the external circuit B1 copper is the positive electrode, where Cu²⁺ is reduced, and E°cell = 0.34 − (−0.76) = +1.10 V comes out positive only with the cell written Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), zinc on the left

Question 18

Identify the positive electrode of the cell Mg(s) | Mg²⁺(aq) || Cu²⁺(aq) | Cu(s), and the direction in which electrons flow in the external circuit.

Accepted Answer

B1 copper is the positive electrode, as the right-hand half-cell with the more positive E° B1 electrons flow through the external wire from the magnesium to the copper

Question 19

Suggest two reasons why a student's zinc–copper cell reads 0.92 V when the expected EMF is 1.10 V.

Accepted Answer

B1 an oxide layer or grease on a metal strip, so the equilibrium at its surface is not the one intended B1 a voltmeter of too low a resistance, so a current flows and the reading falls below the EMF; or a dry or contaminated salt bridge

Cell, copper on the right	Reading / V
Nickel–copper	0.57
Lead–copper	0.46
Zinc–copper	1.09

Half-equation	E° / V
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s)	−2.37
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)	−0.76
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s)	+0.34
Ag⁺(aq) + e⁻ ⇌ Ag(s)	+0.80