Electrode potentials are measured against the standard hydrogen electrode.

The standard hydrogen electrode 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 and 2 only is the correct answer.

The species discharged at the cathode/negative electrode (gaining e^–) will be lower (less –ve) in the electrochemical series; water is lower than potassium so water is discharged, forming hydrogen gas (hydrogen is lower in the activity series than potassium). Without the data booklet we must remember that the least reactive metal/hydrogen is discharged.

The species discharged at the anode/positive electrode (losing e^–) will be higher (less +ve) in the electrochemical series. Again, without the data booklet we must remember that the least reactive non-metal is discharged. 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 Cathode: H₂ Anode: Cl₂ is the correct answer.

During electrolysis of a molten metal halide, the metal ions will be discharged at the cathode (negative electrode) and the halide ions will be discharged at the anode (positive electrode).

The amount of charge passed can be found thus: charge (coulombs, C) = current (amps, A) × time (seconds, s), so both current and time will affect the amount of electrons and therefore the amount of product formed at the anode and cathode.

However, the metal ion is discharged at the cathode, rather than the anode, so the charge on the metal ion will have no impact on the amount of product at the anode.

Thus 1 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).

The cell potential is the difference (as on a number line) in the potential of the two half-cells.

Or mathematically E^ocell = E^ocathode – E^oanode = +0.34 – –1.18 = +1.52 V

Thus +1.52 V is the correct answer.

If electrodes are inert (e.g. platinum) then they do not get involved in the electrolysis process. Electrodes that are made of other metals may be preferentially involved in the electrolysis, depending on position in the electrochemical series.

If nickel electrodes (instead of inert electrodes) are used to electrolyse nickel sulfate (NiSO₄(aq)) solution then nickel will be lost from the anode (positive electrode) and gained at the cathode (negative electrode).

Therefore the object needs to be placed as the cathode (Y) as nickel is deposited on the cathode, and the anode (X) needs to be nickel, as nickel is lost from the anode (instead of oxygen being produced). The solution needs to be a nickel salt solution, in this case NiSO₄(aq).

Thus Electrode X: Nickel Electrode Y: Object Electrolyte Z: NiSO₄(aq) is the correct answer.

Moles of gallium = mass/Ar = \(x \over 70\)

Half equations: Ga³⁺ (aq) + 3e^– → Ga (s) and Mg²⁺ (aq) + 2e^– → Mg (s) so ratio of moles of electrons for gallium to moles of electrons for magnesium is 3:2.

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

Moles of magnesium = \(x \over 70\) × \(3 \over 2\)

Mass of magnesium = moles × Ar = \(x \over 70\) × \(3 \over 2\) × 24

Which multiplied out gives \(72x \over 140\) which is therefore the correct answer.

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.

Therefore the iron half-cell must be more negative (E^o) than the silver half-cell (since the iron loses electrons, which is oxidation) and the silver half-cell which gains electrons (reduction) is more positive (E^o). Electrons will flow from negative to positive; from iron to silver.

Thus 1, 2 and 3 is the correct answer.

For any reaction to be spontaneous Gibbs Free Energy, ΔG^o, must be negative.

And since ΔG^o = −nFE^o_cell (this should be remembered) the standard cell potential, E^o_cell , will always be positive for a spontaneous reaction.

Therefore, E^o_cell must be positive and ΔG^o must be negative is the correct answer.

The species discharged at the cathode/negative electrode (gaining e^–) will be lower (less –ve) in the electrochemical series. Without the data booklet we must remember that the least reactive metal/hydrogen is discharged.

Water is lower than potassium so water is discharged preferentially to potassium, forming hydrogen gas (hydrogen is lower in the activity series than potassium). Copper is lower than water so copper will be preferentially discharged over water.

The species discharged at the anode/positive electrode (losing e^–) will be higher (less +ve) in the electrochemical series. Again, without the data booklet we must remember that the least reactive non-metal is discharged.

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).

Water is preferentially discharged over compound ions (like SO₄, NO₃, etc.) at the anode/positive electrode, forming oxygen gas.

Thus K₂SO₄ (aq) is the correct answer, since water will be preferentially discharged at both cathode and anode, producing hydrogen and oxygen.

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).

Red Cat: Reduction at cathode, An Ox: Oxidation at anode.

The cell potential is the difference (as on a number line) in the potential of the two half-cells.

Or mathematically E^ocell = E^ocathode – E^oanode

Therefore, E^oanode = E^ocathode – E^ocell = +0.34 – +048 = −0.14 V

Thus −0.14 V is the correct answer.