Electrolysis & Metals

7 methods · 18 questions

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In the electrolysis of an aqueous solution with inert electrodes, water competes with the solute at both electrodes, so the product is decided by two fixed rules rather than by the salt's name.

Trigger: the words 'aqueous', 'solution', 'dilute' or 'concentrated' next to an electrolysis.

Instances:
(i) at the cathode, a metal is deposited only if it is below hydrogen in the reactivity series ( $Cu$ , $Ag$ , $Pb$ ); a metal above hydrogen ( $Na$ , $K$ , $Ca$ , $Mg$ , $Al$ ) is never discharged and hydrogen gas comes off instead;
(ii) at the anode, a concentrated halide solution gives the halogen ( $C l_{2}$ , $B r_{2}$ ); with no halide (or a very dilute one) oxygen comes off from the water;
(iii) $S O_{4}^{2 -}$ and $N O_{3}^{-}$ are spectators, never discharged, so they always leave oxygen at the anode.

Linked questions (3)

NSAA 2016 Section 1, Part C (Chemistry questions) · Q47

During the electrolysis of a saturated solution of sodium chloride,

2.4 dm^{3}

of hydrogen gas was collected in time

t

at one of the electrodes.

Assuming no products dissolve, which row in the table correctly gives the mass or volume of the given product collected at the given electrode in time

t

?

(

A_{r}

: Na = 23; Cl = 35.5; H = 1, 1 mole of gas occupies

24 dm^{3}

at room temperature and pressure)

The options list, in order: the mass or volume, the product and the electrode.

A $0.1 g$ ; hydrogen; negative
B $2.3 g$ ; sodium; negative
C $2.4 dm^{3}$ ; chlorine; positive
D $2.4 dm^{3}$ ; chlorine; negative
E $2.4 dm^{3}$ ; oxygen; positive
F $3.55 g$ ; chlorine; positive
G $1.2 dm^{3}$ ; oxygen; negative

Worked solution

Answer:C

In the electrolysis of saturated sodium chloride solution, hydrogen is produced at the negative electrode and chlorine at the positive electrode; sodium ions stay in solution and no oxygen is formed.

Cathode:

2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}

. Anode:

2 C l^{-} \to C l_{2} + 2 e^{-}

. Two electrons give one molecule of each gas, so hydrogen and chlorine are produced in a 1 : 1 mole ratio.

2.4 dm^{3}

of hydrogen is

\frac{2.4}{24} = 0.1 mol

, so

0.1 mol

of chlorine is collected at the positive electrode:

0.1 \times 24 = 2.4 dm^{3}

, or

0.1 \times 71 = 7.1 g

.

Row A is wrong because

0.1 mol

of hydrogen has a mass of

0.2 g

; row B because sodium metal is not deposited from aqueous solution; row D because chlorine forms at the positive electrode; rows E and G because no oxygen is produced; row F because the mass of chlorine would be

7.1 g

.

The answer is C.

NSAA 2017 Section 1, Part C (Chemistry questions) · Q45

Consider this electrochemical cell containing an aqueous copper(II) chloride electrolyte:

Which row in the following table identifies the reactions occurring at the electrodes?

The options list, in order: the reaction at the cathode (negative electrode) and the reaction at the anode (positive electrode).

An electrolysis cell drawn as a rectangular beaker containing a liquid labelled 2 mol dm^-3 CuCl2(aq) electrolyte. Two vertical electrodes dip into the liquid and are joined above the beaker by wires to a cell marked 9 V d.c. The left electrode, connected to the terminal marked with a minus sign, is labelled pure copper cathode (negative electrode). The right electrode, connected to the terminal marked with a plus sign, is labelled graphite anode (positive electrode).

A $C u^{2 +} (aq) + 2 e^{-} \to Cu (s)$ ; $Cu (s) \to C u^{2 +} (aq) + 2 e^{-}$
B $Cu (s) \to C u^{2 +} (aq) + 2 e^{-}$ ; $C u^{2 +} (aq) + 2 e^{-} \to Cu (s)$
C $2 C l^{-} (aq) \to C l_{2} (g) + 2 e^{-}$ ; $C u^{2 +} (aq) + 2 e^{-} \to Cu (s)$
D $C u^{2 +} (aq) + 2 e^{-} \to Cu (s)$ ; $2 C l^{-} (aq) \to C l_{2} (g) + 2 e^{-}$
E $Cu (s) \to C u^{2 +} (aq) + 2 e^{-}$ ; $2 O H^{-} (aq) \to H_{2} (g) + O_{2} (g) + 2 e^{-}$

Worked solution

Answer:D

In electrolysis, positive ions move to the negative electrode (cathode) and gain electrons; negative ions move to the positive electrode (anode) and lose electrons.

At the copper cathode, the electrolyte contains

Cu^{2 +}

ions and, from water,

H^{+}

ions. Copper is less reactive than hydrogen, so copper ions are discharged:

C u^{2 +} (aq) + 2 e^{-} \to Cu (s)

. This rules out rows B, C and E, which either oxidise copper at the cathode or discharge chloride there.

At the graphite anode the choice is between

C l^{-}

and

O H^{-}

ions. The anode is inert, so it does not dissolve, which rules out row A. The chloride solution is concentrated (

2 mol dm^{- 3}

), so chloride ions are discharged in preference to hydroxide ions and chlorine gas is released:

2 C l^{-} (aq) \to C l_{2} (g) + 2 e^{-}

.

Row D gives copper deposited at the cathode and chlorine evolved at the anode.

The answer is D.

NSAA 2018 Section 1, Part C (Chemistry questions) · Q39

What is the overall process that takes place at the cathode (negative electrode) in the electrolysis of dilute aqueous sodium sulfate?

A $2 H^{+} + 2 e^{-} \to H_{2}$
B $2 O^{2 -} \to O_{2} + 4 e^{-}$
C $4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}$
D $N a^{+} + e^{-} \to Na$
E $S O_{4}^{2 -} \to S O_{2} + O_{2} + 2 e^{-}$

Worked solution

Answer:A

The cathode is where reduction happens: positive ions gain electrons. The solution contains

N a^{+}

from the salt and

H^{+}

from water. Sodium is far more reactive than hydrogen, so

N a^{+}

stays in solution and it is hydrogen ions that are discharged.

The overall cathode process is therefore

2 H^{+} + 2 e^{-} \to H_{2},

and hydrogen gas is collected at the negative electrode.

Options B, C and E describe oxidations (electrons are released), which can only happen at the anode, and option D would require sodium metal to be deposited from water, which does not occur.

The answer is A.