In a voltaic cell the redox reaction generates the charges on the electrodes and causes the current to flow. The half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) causing the electrode to be negative, and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction) causing the electrode to be positive.

Zinc half-cell is more negative than the iron half-cell (Electrochemical series in the data book) so the zinc half-cell will lose electrons (oxidation) and the iron half-cell will gain electrons (reduction); electrons will flow from the zinc to iron.

E^ocell = E^ocathode – E^oanode = +0.77 – –0.76 = +1.53 V

As electrons are flowing from zinc to iron, the charge will be equalised by negative ions flowing across the salt bridge from the iron to the zinc half-cell.

Thus 2 and 3 only is the correct answer.

In a voltaic cell the half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction).

Nickel half-cell is more negative than the copper half-cell (Electrochemical series in the data book) so the nickel half-cell will lose electrons (oxidation) and the copper half-cell will gain electrons (reduction):

Ni_(s) → Ni²⁺_(aq) + 2e^– and Cu²⁺_(aq) + 2e^– → Cu_(s)

The number of electrons lost and gained is the same so the half-equations combined:

Cu²⁺_(aq) + Ni_(s) ⇌ Ni²⁺_(aq) + Cu_(s)

E^ocell = E^ocathode – E^oanode = +0.34 – –0.26 = +0.60 V

Thus Cu²⁺_(aq) + Ni_(s) ⇌ Ni²⁺_(aq) + Cu_(s) E^ocell = +0.60 V is the correct answer.

In a voltaic cell the half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction).

Aluminium half-cell is more negative than the nickel half-cell (Electrochemical series in the data book) so the aluminium half-cell will lose electrons (oxidation) and the nickel half-cell will gain electrons (reduction):

Al_(s) → Al³⁺_(aq) + 3e^– and Ni²⁺_(aq) + 2e^– → Ni_(s)

The number of electrons lost and gained in each half-cell must be equalised when the half-equations combine (Al × 2, Ni × 3):

3Ni²⁺_(aq) + 2Al_(s) ⇌ 2Al³⁺_(aq) + 3Ni_(s)

E^ocell = E^ocathode – E^oanode = –0.26 – –1.66 = +1.40 V

Note that the E^o values are independent of the multiplication to equal out the number of electrons.

Thus 3Ni²⁺_(aq) + 2Al_(s) ⇌ 2Al³⁺_(aq) + 3Ni_(s) E^ocell = +1.40V is the correct answer.

Electrode potentials are measured against the SHE; standard hydrogen electrode.

The SHE consists of a feed of hydrogen gas at 100 kPa onto an inert Pt electrode in 1 mol dm^–3 hydrochloric acid all at 298 K.

The platinum electrode is coated with platinum black; finely powdered platinum to give a high surface area for the reaction to occur on: H⁺_(aq) + e^– ⇌ ½H_2(g) E^o = 0.00 V

Thus 1, 2 and 3 is the correct answer.

An inert platinum electrode is needed for all gaseous electrodes (e.g. chlorine and hydrogen) and for half-cells involving only aqueous ions (e.g. Cu²⁺_(aq) + e^– ⇌ Cu⁺_(aq)).

The copper half-cell Cu²⁺_(aq) + 2e^– ⇌ Cu_(s) requires a copper electrode and is therefore the correct answer.

Iron(II) half-cell is more negative than the copper half-cell (Electrochemical series in the data book) so the iron half-cell will lose electrons (oxidation) and the copper half-cell will gain electrons (reduction):

Fe_(s) → Fe²⁺_(aq) + 2e^– and Cu²⁺_(aq) + 2e^– → Cu_(s)

The number of electrons lost and gained is the same (that is two electrons) so the half-equations combined:

Cu²⁺_(aq) + Fe_(s) ⇌ Fe²⁺_(aq) + Cu_(s)

E^ocell = E^ocathode – E^oanode = +0.34 – –0.45 = +0.79 V

ΔG^o = –nFE^o_cell = – (2 mol × 96500 C mol^–1 × +0.79 V) = – 152470 J = –152 kJ (3 sig figs)

Incorrect answers

–76.2 uses 1 mol of electrons instead of 2

+152 uses –0.79 V instead of +0.79 V (or emits to use the –ve sign in the equation)

+76.2 uses 1 mol of electrons instead of 2 and uses –0.79 V instead of +0.79 V (or emits to use the –ve sign in the equation)

The species discharged at the negative electrode, gaining e^– will be lower (less –ve) in the electrochemical series; copper is lower than water so copper ions are discharged.

Normally the species discharged at the positive electrode, losing e^–, will be higher (less +ve) in the electrochemical series. Note! Exception: Chloride ions (½Cl_2(g) + e^– ⇌ Cl^–_(aq) E^o = +1.36 V) are preferentially discharged over water (½O_2(g) + 2H⁺_(aq) + 2e^– ⇌ H₂O_(l) E^o = +1.23 V) at the positive electrode despite their E^o values (this is a kinetic effect that needs remembering).

However, if a copper electrode is used instead of an inert electrode, copper from the electrode will lose electrons preferentially over chloride ions or water (Cu²⁺_(aq) + 2e^– ⇌ Cu_(s) E^o = +0.34 V) because its E^o is higher in the electrochemical series.

Thus negative: Cu²⁺_(aq) + 2e^– → Cu_(s) positive: Cu_(s) → Cu²⁺_(aq) + 2e^– is the correct answer.

The species discharged at the negative electrode, gaining e^– will be lower (less –ve) in the electrochemical series; copper is lower than water so copper ions are discharged.

The species discharged at the positive electrode, losing e^–, will be higher (less +ve) in the electrochemical series. Note! Exception: Chloride ions (½Cl_2(g) + e^– ⇌ Cl^–_(aq) E^o = +1.36 V) are preferentially discharged over water (½O_2(g) + 2H⁺_(aq) + 2e^– ⇌ H₂O_(l) E^o = +1.23 V) at the positive electrode despite their E^o values (this is a kinetic effect that needs remembering).

Thus copper and chlorine is the correct answer.

Charge = current × time = 5.0 × (30.0 × 60) = 9000C

Moles of electrons passed through circuit = charge/Faraday = 9000/96500 = 0.0932642

Half equation: Zn²⁺_(aq) + 2e^– → Zn(s) so mole ratio of electrons to zinc atoms is 2:1,

Therefore mass of zinc = moles × Ar = 0.0932642/2 × 65.38 = 3.0488083 = 3.05g (3sf)

Therefore 3.05 is the correct answer.

Incorrect answers

6.10 uses a 1:1 mole ratio for electrons to atoms instead of 2:1

0.0508 uses 150C of charge (forgetting to multiply minutes by 60 to get seconds)

0.102 uses 150C of charge (forgetting to multiply minutes by 60 to get seconds) and uses a 1:1 mole ratio for electrons to atoms instead of 2:1

Moles of aluminium = mass/Ar = 1.62/26.98 = 0.0600444 mol

Half equations: Al³⁺_(aq) + 3e^– → Al_(s) and Ca²⁺_(aq) + 2e^– → Ca_(s) so ratio of moles of electrons for aluminium to moles of electrons for calcium is 3:2.

This means that the ratio of moles of aluminium:calcium will be 2:3 (since the same number of electrons will pass through the circuit, but that will give more calcium atoms as only two electrons are needed for each atom, rather than 3 for aluminium).

Moles of calcium = 0.0600444 × 3/2 = 0.0900666

Mass of calcium = moles × Ar = 0.0900666 × 40.08 = 3.6098693 = 3.61g (3sf)

Therefore 3.61 is the correct answer.

Incorrect answers

1.60 uses a 3:2 mole ratio for moles of aluminium:calcium instead of 2:3

2.41 uses a 1:1 mole ratio for moles of aluminium:calcium instead of 2:3

4.81 uses a 1:2 mole ratio for moles of aluminium:calcium instead of 2:3