Why does aqueous NaCl give hydrogen at cathode instead of sodium?

The reduction potential of Na+/Na is -2.71 V, while water is reduced at about -0.41 V at neutral pH (-0.83 V at unit hydroxide). Water is far easier to reduce, so hydrogen is liberated. Sodium is deposited only from molten NaCl where no water is present.

Why is current efficiency less than 100% in industrial electrolysis?

Some current is lost to side reactions, resistive heating, and back-reactions. Industrial cells typically achieve 85-95% efficiency.

What is the role of overvoltage in electrolysis?

Overvoltage is the extra voltage needed to actually discharge gases like H2 and O2 beyond the theoretical value. It depends on electrode material and current density.

Electrolytic Cells And Electrolysis

Q: What is the difference between electrolytic and electrochemical cells?

An electrolytic cell consumes electrical energy from an external source to drive a non-spontaneous reaction. An electrochemical (galvanic) cell produces electrical energy from a spontaneous redox reaction. Cathode is negative in an electrolytic cell but positive in a galvanic cell.

Q: What is 1 Faraday and why is it 96500 C?

1 Faraday is the charge on 1 mole of electrons. With each electron carrying 1.602e-19 C and 1 mole containing 6.022e23 electrons, the product is 96485 C, rounded to 96500 C.

Q: Can Faraday's laws be applied to molten electrolytes?

Yes. Faraday's laws apply universally to any electrolysis, whether the electrolyte is molten or in aqueous solution.

Q: What decides whether an anode is inert or reactive?

If the electrode metal is more easily oxidised than the anion in solution, the electrode dissolves (Cu, Ag, Zn are reactive). Pt, Au, graphite are inert and let the anion or water get oxidised instead.

Q: How is aluminium extracted commercially?

The Hall-Heroult process electrolyses molten Al2O3 dissolved in molten cryolite at about 950 degrees Celsius. Cryolite lowers the melting point and improves conductivity.

ChemistryElectrochemistryFor NEET aspirants

An electrolytic cell uses external electrical energy to drive a non-spontaneous redox reaction, splitting an electrolyte (molten salt or aqueous solution) into its ions and depositing them at electrodes. Electrolysis is governed by Faraday's two laws, which quantitatively link the mass of substance deposited to the charge passed. The concept underpins electroplating, metal refining, industrial production of $Cl_{2}$ , $NaOH$ , aluminium, and every JEE/NEET numerical on "how much metal deposits when $I$ amperes flow for $t$ seconds."

Key Formulas - Quick Reference

Charge passed: $Q = I \times t$ (coulombs, when $I$ is in amperes and $t$ in seconds)
1 Faraday $= 96500 C =$ charge on 1 mole of electrons
Faraday's 1st law: $w = Z \cdot I \cdot t = Z \cdot Q$
Electrochemical equivalent: $Z = \frac{E}{96500}$ where $E = \frac{M}{n}$ (equivalent weight)
Faraday's 2nd law: $\frac{w _{1}}{w _{2}} = \frac{E _{1}}{E _{2}}$ (same charge through different cells)
Moles deposited: $n_{deposit} = \frac{I \cdot t}{n \cdot F}$ where $n$ = electrons per ion
Current efficiency: $η = \frac{actual mass deposited}{theoretical mass} \times 100%$
Gas volume at STP: $V = \frac{Q}{n F} \times 22400$ mL (per mole of gas)

1. Ionic Theory and Electrolytes

The Arrhenius ionic theory proposes that when an electrolyte dissolves in water (or is melted), it dissociates into positive and negative ions. These free ions carry current in solution, unlike metals where free electrons carry current.

1.1 Strong vs Weak Electrolytes

Type	Behaviour	Examples
Strong	Nearly completely dissociated ( $α \approx 1$ ); ion concentration $≫$ molecular concentration	$NaCl$ , $KOH$ , $HCl$ , $HNO_{3}$ , $H_{2} SO_{4}$ , $NaOH$
Weak	Partially dissociated ( $α ≪ 1$ ); molecular form dominates	$CH_{3} COOH$ , $NH_{4} OH$ , $HCN$ , $H_{2} CO_{3}$

Degree of dissociation

(α) = \frac{moles of dissociated molecules}{total moles taken}

, with range

0 < α < 1

1.2 Factors Affecting Degree of Dissociation

Nature of solute: Ionic compounds dissociate more than covalent.
Nature of solvent: High dielectric constant solvents (like water, $ε = 80$ ) promote dissociation.
Dilution: $α$ increases with dilution (Ostwald's dilution law: $α \propto \frac{1}{C}$ for weak electrolytes).
Temperature: Generally increases with temperature.
Common ion effect: Presence of a common ion suppresses dissociation.

2. Electrolysis: The Setup

Electrolysis is the decomposition of an electrolyte by passing direct current through its molten state or aqueous solution. An external battery pumps electrons from the anode to the cathode via the external circuit; inside the electrolyte, ions migrate to complete the circuit.

Cathode (connected to negative terminal of battery): electrons enter here; reduction occurs (cations gain electrons).
Anode (connected to positive terminal): electrons leave here; oxidation occurs (anions lose electrons).

Figure 1: An electrolytic cell. The battery does the work, forcing electrons into the cathode and dragging them off the anode so a non-spontaneous redox reaction is driven forward. The signs are the reverse of a galvanic cell, but the roles are not: the cathode is still where reduction happens.

2.1 Ionisation vs Electrolysis

Ionisation happens the moment the electrolyte dissolves (no current needed). Electrolysis is the subsequent discharge of ions at electrodes when current flows. So ions exist first; current only decides where they go and what they become.

3. Preferential Discharge Theory

When more than one cation (or anion) is present, only one gets discharged first at each electrode. The selection depends on:

Position in electrochemical series (standard reduction potential): ions with higher reduction potential are discharged first at the cathode; ions with lower reduction potential (more easily oxidised) are discharged first at the anode.
Concentration: more concentrated ion is preferred (mass-action effect).
Nature of electrode: some electrodes participate (e.g., Cu anode in $CuSO_{4}$ solution) and dissolve instead of oxidising water.
Overvoltage: practical extra voltage needed beyond theoretical; for gases like $O_{2}$ and $H_{2}$ , overvoltage on certain electrodes changes the outcome.

3.1 Discharge Order (approximate, from easy to difficult)

At Cathode (reduction)	At Anode (oxidation)
Easier: $Au^{3 +} > Ag^{+} > Cu^{2 +} > H^{+} > Pb^{2 +} > Fe^{2 +} > Zn^{2 +} > Al^{3 +} > Mg^{2 +} > Na^{+} > K^{+}$	Easier: $OH^{-} > I^{-} > Br^{-} > Cl^{-} > F^{-} > NO_{3}^{-} > SO_{4}^{2 -}$

Quick rule: If the anion is

SO_{4}^{2 -}

NO_{3}^{-}

, or

F^{-}

(all hard to oxidise), water is oxidised at the anode instead, releasing

O_{2}

. If the cation is

Na^{+}

K^{+}

Ca^{2 +}

etc. (all hard to reduce), water is reduced at the cathode instead, releasing

H_{2}

3.2 Common Examples

Electrolyte	At Cathode	At Anode
Molten $NaCl$	$Na$ deposited	$Cl_{2}$ gas
Aqueous $NaCl$ (brine)	$H_{2}$ gas ( $Na^{+}$ too hard)	$Cl_{2}$ gas (overvoltage of $O_{2}$ )
Aqueous $CuSO_{4}$ (Pt electrodes)	$Cu$ deposited	$O_{2}$ gas ( $SO_{4}^{2 -}$ inert)
Aqueous $CuSO_{4}$ (Cu electrodes)	$Cu$ deposited	$Cu$ anode dissolves
Dilute $H_{2} SO_{4}$	$H_{2}$ gas	$O_{2}$ gas
Aqueous $AgNO_{3}$ (Pt)	$Ag$ deposited	$O_{2}$ gas

Figure 2: Same salt, different products. Melting NaCl removes the only species that can outcompete Na⁺ for electrons, which is why sodium is manufactured from the melt (Down's process) and never from brine.

Solved Example 1

The pH of a solution of

NaCl

is 7. This solution is electrolysed using Pt electrodes. Predict the pH after passage of some current.

Solution:

At cathode: $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ ( $Na^{+}$ not discharged).

At anode: $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ ( $Cl^{-}$ discharged over $OH^{-}$ due to overvoltage effect on Pt).

Net: solution accumulates $OH^{-}$ , so pH rises above 7 (becomes alkaline).

4. Faraday's Laws of Electrolysis

4.1 First Law

The mass (

w

) of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (

Q

) passed through the electrolyte.

w \propto Q ⟹ w = Z \cdot Q = Z \cdot I \cdot t

where

Z

is the electrochemical equivalent (mass deposited per coulomb).

4.2 Second Law

When the same quantity of electricity is passed through different electrolytes in series, the masses deposited are in the ratio of their equivalent weights.

\frac{w _{1}}{w _{2}} = \frac{E _{1}}{E _{2}}

Figure 3: Faraday's second law made visible. One loop means one value of

Q

, so the three deposits differ only through their equivalent weights

E = M / n

. Blue outline marks the cathode, where the metal plates out; orange marks the anode.

4.3 Electrochemical Equivalent (Z)

$Z$ is the mass deposited by 1 coulomb of charge. Since 1 mole of electrons ( $= 1$ Faraday $= 96500$ C) deposits 1 gram-equivalent:

Z = \frac{E}{F} = \frac{E}{96500} = \frac{M}{n \cdot 96500}

where $M$ = molar mass and $n$ = number of electrons per ion (valence).

4.4 Gram Equivalent Weight

Equivalent weight $E = \frac{M}{n}$ where $n$ = number of electrons transferred per formula unit in the electrode reaction. For example:

$Ag^{+} + e^{-} \to Ag$ : $n = 1$ , $E = 108$
$Cu^{2 +} + 2 e^{-} \to Cu$ : $n = 2$ , $E = 63.5/2 = 31.75$
$Al^{3 +} + 3 e^{-} \to Al$ : $n = 3$ , $E = 27/3 = 9$
$2 H^{+} + 2 e^{-} \to H_{2}$ : $n = 2$ for 1 mole $H_{2}$ , $E = 1$ (for atomic H)

Molar interpretation: Moles of substance deposited

= \frac{Q}{n F} = \frac{I \cdot t}{n \cdot 96500}

. This form is easier for gases (compute moles, then convert to volume).

Solved Example 2

A vanadium electrode is oxidised electrically. If the mass of the electrode decreases by 0.4523 g while 2570 coulombs pass through the external circuit, what is the oxidation state of vanadium in the product? (At. wt. V = 50.94)

Solution:

Moles of $V$ oxidised $= \frac{0.4523}{50.94} = 8.879 \times 1 0^{- 3}$ mol.

Moles of electrons passed $= \frac{Q}{F} = \frac{2570}{96500} = 2.663 \times 1 0^{- 2}$ mol.

Electrons lost per vanadium atom $= \frac{2.663 \times 1 0 ^{- 2}}{8.879 \times 1 0 ^{- 3}} = 3.00$ .

Each V atom gives up 3 electrons, so the oxidation state in the product is +3.

Method: whenever a question hands you a mass and a charge, convert both to moles and divide. That ratio is

n

, and it must land on a whole number. If it does not, the data is inconsistent rather than the chemistry.

Solved Example 3

For how long must a current of 3 A be passed through a

CuSO_{4}

solution to deposit 3 g of copper? (At. wt. Cu = 63.5)

Solution:

$Cu^{2 +} + 2 e^{-} \to Cu$ ; equivalent weight $E = 63.5/2 = 31.75$ g/equivalent.

By Faraday's 1st law: $w = \frac{E \cdot I \cdot t}{96500}$ , so $t = \frac{w \cdot 96500}{E \cdot I} = \frac{3 \times 96500}{31.75 \times 3} \approx 3039$ s $\approx 50.6$ minutes.

Solved Example 4

Find the charge in coulombs on 1 gram-ion of

N^{3 -}

Solution:

1 gram-ion contains $N_{A}$ ions. Each $N^{3 -}$ carries a charge of $3 e = 3 \times 1.6 \times 1 0^{- 19}$ C.

Total charge $= 3 \times 1.6 \times 1 0^{- 19} \times 6.022 \times 1 0^{23} = 2.89 \times 1 0^{5}$ C $= 3 F$ (since $1 F = 96500$ C, $3 F = 289500$ C).

Solved Example 5

How many moles of iron will be produced by passage of 4 A of current through molten

FeCl_{3}

for 10 minutes?

Solution:

$Q = I \cdot t = 4 \times 600 = 2400$ C. $Fe^{3 +} + 3 e^{-} \to Fe$ ; so moles of Fe $= \frac{Q}{n F} = \frac{2400}{3 \times 96500} = 8.29 \times 1 0^{- 3}$ mol.

Solved Example 6

The same amount of electricity was passed through molten cryolite (

Na_{3} AlF_{6}

) and a

ZnSO_{4}

solution. If 1.8 g of Al is deposited, how much Zn is deposited? (At. wt: Al = 27, Zn = 65.4)

Solution:

By 2nd law: $\frac{w _{Zn}}{w _{Al}} = \frac{E _{Zn}}{E _{Al}} = \frac{65.4/2}{27/3} = \frac{32.7}{9}$ .

$w_{Zn} = 1.8 \times \frac{32.7}{9} = 6.54$ g.

Solved Example 7

Find the volume of gas (at STP: 273 K, 1 atm) liberated at the anode when dilute

H_{2} SO_{4}

is electrolysed with 0.965 A for 1 hour.

Solution:

At anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ ; so 4 moles of electrons liberate 1 mole $O_{2}$ .

$Q = 0.965 \times 3600 = 3474$ C. Moles $e^{-} = 3474/96500 = 0.036$ . Moles $O_{2} = 0.036/4 = 0.009$ .

Volume at STP $= 0.009 \times 22400 = 201.6$ mL.

Solved Example 8

A current of 0.20 A is passed for 482 s through 50.0 mL of 0.100 M

KI

solution. Assume

I_{2}

liberated at the anode reacts fully with

I^{-}

to form

I_{3}^{-}

. Find the concentration of

I_{3}^{-}

formed.

Solution:

$Q = 0.20 \times 482 = 96.4$ C; moles $e^{-} = 96.4/96500 = 1 0^{- 3}$ .

At anode: $2 I^{-} \to I_{2} + 2 e^{-}$ ; moles $I_{2} = 1 0^{- 3} /2 = 5 \times 1 0^{- 4}$ .

Each $I_{2}$ combines with 1 $I^{-}$ : $I_{2} + I^{-} \to I_{3}^{-}$ , so moles $I_{3}^{-} = 5 \times 1 0^{- 4}$ .

Concentration $= 5 \times 1 0^{- 4} /0.050 = 1 0^{- 2}$ M $= 0.01$ M.

5. Current Efficiency

In practical electrolysis, not all the current does useful work; some is wasted on side reactions (e.g., water electrolysis, resistive heating, secondary reactions). Current efficiency quantifies this:

η = \frac{actual mass deposited}{theoretical mass by Faraday’s law} \times 100%

Solved Example 9

A solution containing 1 mol/L each of

Cu(NO_{3})_{2}

AgNO_{3}

Hg_{2} (NO_{3})_{2}

, and

Mg(NO_{3})_{2}

is electrolysed using inert electrodes. Order in which the metals deposit at the cathode?

Solution:

Higher reduction potential deposits first: $E ° (Ag^{+} / Ag) = 0.80$ , $E ° (Hg_{2}^{2 +} / Hg) = 0.79$ , $E ° (Cu^{2 +} / Cu) = 0.34$ , $E ° (Mg^{2 +} / Mg) = - 2.37$ V.

Order: $Ag > Hg > Cu$ , and $Mg$ is never deposited (water reduces to $H_{2}$ instead).

Solved Example 10

One coulomb of charge passes through a solution of

AgNO_{3}

and

CuSO_{4}

connected in series. Find the ratio of masses of Ag to Cu deposited.

Solution:

By 2nd law: $\frac{w _{Ag}}{w _{Cu}} = \frac{E _{Ag}}{E _{Cu}} = \frac{108/1}{63.5/2} = \frac{108}{31.75} = 3.40$ .

Solved Example 11

0.3605 g of a metal is deposited on the electrode by passing 1.2 A for 15 minutes through its solution. If the atomic weight of the metal is 96, calculate its valency.

Solution:

$Q = 1.2 \times 900 = 1080$ C. Moles $e^{-} = 1080/96500 = 0.01119$ .

Moles of metal $= 0.3605/96 = 3.755 \times 1 0^{- 3}$ .

Valency $n = 0.01119/3.755 \times 1 0^{- 3} \approx 2.98 \approx 3$ .

Solved Example 12

A current of 4 A was passed for 1.5 hours through a solution of

CuSO_{4}

; 3.0 g of

Cu

was deposited. Find the current efficiency. (At. wt. Cu = 63.5)

Solution:

Theoretical mass: $w = \frac{E \cdot I \cdot t}{96500} = \frac{31.75 \times 4 \times 5400}{96500} = 7.108$ g.

Efficiency $= \frac{3.0}{7.108} \times 100 = 42.2%$ .

Solved Example 13

How many grams of silver can be plated out on a serving tray by electrolysis of a solution containing

Ag^{+}

ions for 8 hours at a current of 8.46 A?

Solution:

$Q = 8.46 \times 8 \times 3600 = 243, 648$ C. Moles Ag $= Q / (n F) = 243648/96500 = 2.525$ mol.

Mass $= 2.525 \times 108 = 272.7$ g.

Solved Example 14

A current of 40 microamperes is passed through a solution of

AgNO_{3}

for 32 minutes using Pt electrodes. What mass of Ag will be deposited?

Solution:

$Q = 40 \times 1 0^{- 6} \times 32 \times 60 = 0.0768$ C.

Mass Ag $= \frac{E \cdot Q}{96500} = \frac{108 \times 0.0768}{96500} = 8.6 \times 1 0^{- 5}$ g $= 86 μ$ g.

6. Industrial Applications of Electrolysis

Electroplating: depositing a thin metal layer (Ag, Cr, Ni) on cheaper base metals for decoration or corrosion protection. The object is the cathode; the plating metal is the anode.
Electrorefining: impure metal is the anode; pure metal deposits on the cathode. Used for Cu, Ag, Au, Zn.
Extraction of reactive metals: Na (from molten $NaCl$ , Down's process), Al (from molten $Al_{2} O_{3}$ in cryolite, Hall-Heroult process), Mg from $MgCl_{2}$ .
Chlor-alkali industry: electrolysis of brine gives $Cl_{2}$ , $H_{2}$ , and $NaOH$ .
Anodising: forming a protective oxide layer on Al by making it the anode in dilute $H_{2} SO_{4}$ .

Figure 4: Electroplating. The article being coated is always the cathode; a bar of the coating metal is the anode and dissolves at roughly the same rate as metal plates out, so the bath never runs down. Only the submerged part gets plated.

Common Mistakes to Avoid

Watch out

Confusing anode/cathode signs: in an electrolytic cell, cathode is negative (opposite of galvanic cell where cathode is positive).
Forgetting to divide molar mass by $n$ (electrons per ion) when computing equivalent weight.
Using $n = 1$ for $Cu^{2 +}$ or $n = 2$ for $Al^{3 +}$ : always check the electrode half-reaction.
Assuming $Na$ or $Al$ deposits from aqueous solution: they don't; water reduces first.
Forgetting that with a Cu anode in $CuSO_{4}$ , the anode dissolves ( $Cu \to Cu^{2 +} + 2 e^{-}$ ) instead of releasing $O_{2}$ .
Mixing units: $Q$ in coulombs $=$ (amperes) $\times$ (seconds). Convert minutes and hours before multiplying.
Using $22400$ mL for gas at 25°C instead of 0°C (STP). At NTP (25°C, 1 atm), 1 mole $= 24, 450$ mL; be strict about the temperature specified.

Frequently Asked Questions

Q1. What is the difference between electrolytic and electrochemical cells?

An electrolytic cell consumes electrical energy from an external source to drive a non-spontaneous reaction ( $Δ G > 0$ ). An electrochemical (galvanic) cell produces electrical energy from a spontaneous redox reaction ( $Δ G < 0$ ). Also, cathode is negative in an electrolytic cell but positive in a galvanic cell.

Q2. Why does aqueous $NaCl$ give $H_{2}$ at cathode instead of $Na$ ?

The reduction potential of $Na^{+} / Na$ is $- 2.71$ V. For $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ the standard value is $- 0.83$ V (at $[OH^{-}] = 1$ M), which works out to about $- 0.41$ V at neutral pH. Either way, water is far easier to reduce, so $H_{2}$ is liberated. $Na$ can only be deposited from molten $NaCl$ where no water is present (Down's process).

Q3. What is 1 Faraday and why is it 96500 C?

1 Faraday is the charge on 1 mole of electrons. Each electron has $1.602 \times 1 0^{- 19}$ C, and 1 mole has $6.022 \times 1 0^{23}$ electrons. Multiplying gives $1.602 \times 1 0^{- 19} \times 6.022 \times 1 0^{23} = 96, 485$ C, rounded to 96500 C for calculations.

Q4. Can Faraday's laws be applied to molten electrolytes?

Yes. The laws apply universally to any electrolysis, whether the electrolyte is molten or in solution. For molten electrolytes, the water complication is absent, so metal deposition is direct (e.g., Na from molten $NaCl$ ).

Q5. Why is current efficiency less than 100% in most industrial electrolysis?

Some current is lost to side reactions (electrolysis of water, reduction of dissolved $O_{2}$ ), some to resistive heating (Joule heating), and some to back-reactions at the electrode. Industrial cells often achieve 85 to 95% efficiency; the rest is engineering loss.

Q6. What decides whether an anode is inert or reactive?

If the electrode metal has a lower oxidation potential than the anion in solution, the electrode itself dissolves (reactive: Cu, Ag, Zn). If the electrode is more resistant to oxidation than the anion (Pt, Au, graphite), the anion or water is oxidised instead.

Q7. How is aluminium extracted commercially given that $Al^{3 +}$ can't be reduced from aqueous solution?

The Hall-Heroult process electrolyses molten $Al_{2} O_{3}$ dissolved in molten cryolite ( $Na_{3} AlF_{6}$ ) at around 950°C. Cryolite lowers the melting point of alumina (from 2045°C to about 950°C) and improves conductivity. Molten Al collects at the carbon cathode.

Q8. What is the role of overvoltage?

Overvoltage is the extra voltage (beyond the theoretical cell voltage) needed to actually discharge a species, especially gases like $H_{2}$ and $O_{2}$ . It depends on electrode material and current density. For example, $H_{2}$ overvoltage on Hg is very high (about 1 V), which is why Hg cathodes selectively reduce $Na^{+}$ (not $H^{+}$ ) in the Castner-Kellner process for NaOH.

Previous year questions on Electrolytic Cells And Electrolysis

9 questions from past papers, each with a step-by-step solution.

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1. Ionic Theory and Electrolytes

1.1 Strong vs Weak Electrolytes

1.2 Factors Affecting Degree of Dissociation

2. Electrolysis: The Setup

2.1 Ionisation vs Electrolysis

3. Preferential Discharge Theory

3.1 Discharge Order (approximate, from easy to difficult)

3.2 Common Examples

4. Faraday's Laws of Electrolysis

4.1 First Law

4.2 Second Law

4.3 Electrochemical Equivalent (Z)

4.4 Gram Equivalent Weight

5. Current Efficiency

6. Industrial Applications of Electrolysis

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between electrolytic and electrochemical cells?

Q2. Why does aqueous NaCl give H2​ at cathode instead of Na?

Q3. What is 1 Faraday and why is it 96500 C?

Q4. Can Faraday's laws be applied to molten electrolytes?

Q5. Why is current efficiency less than 100% in most industrial electrolysis?

Q6. What decides whether an anode is inert or reactive?

Q7. How is aluminium extracted commercially given that Al3+ can't be reduced from aqueous solution?

Q8. What is the role of overvoltage?

Previous year questions on Electrolytic Cells And Electrolysis

Ready to master Electrochemistry?

Q2. Why does aqueous $NaCl$ give $H_{2}$ at cathode instead of $Na$ ?

Q7. How is aluminium extracted commercially given that $Al^{3 +}$ can't be reduced from aqueous solution?