Q.A current of $2\ \text{A}$ is passed through molten copper(II) sulfate for $1930\ \text{s}$. Calculate the mass of copper deposited at the cathode. (Atomic mass of $\text{Cu} = 63.5$; $F = 96500\ \text{C mol}^{-1}$.)

Why this approach works — the core idea

Electrolysis is about forcing a non-spontaneous redox reaction using electrical energy. At the cathode (negative electrode), reduction happens — species gain electrons. At the anode (positive electrode), oxidation happens — species lose electrons.

The tricky part with aqueous solutions is that water itself can compete with the dissolved ions. You have to compare the standard reduction potentials of all possible reactions to decide which one actually occurs at each electrode. The rule is simple: at the cathode, the species with the higher (more positive) reduction potential gets reduced first. At the anode, the species with the lower (more negative) reduction potential gets oxidised first (or equivalently, the one that is easiest to oxidise).

For aqueous $CuCl_2$, the ions present are $Cu^{2+}$ and $Cl^-$, plus $H^+$ and $OH^-$ from water's autoionisation.

Step-by-step reasoning

1. What happens at the cathode?

The cathode is negative, so it attracts positive ions ($Cu^{2+}$ and $H^+$ from water). Two reduction reactions are possible:

• $Cu^{2+} + 2e^- \rightarrow Cu(s)$ with $E^\circ = +0.34\ \text{V}$
• $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$ with $E^\circ = -0.83\ \text{V}$

The copper reduction has a much higher (more positive) reduction potential. That means $Cu^{2+}$ is far more willing to accept electrons than water is. So copper metal deposits on the cathode, and no hydrogen gas is produced.

2. What happens at the anode?

The anode is positive, so it attracts negative ions ($Cl^-$ and $OH^-$ from water). Two oxidation reactions are possible:

• $2Cl^- \rightarrow Cl_2(g) + 2e^-$ — this is the reverse of the $Cl_2/Cl^-$ reduction, so its oxidation potential is $-1.36\ \text{V}$ (the negative of the reduction potential).
• $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$ — this is the reverse of the $O_2/H_2O$ reduction, so its oxidation potential is $-1.23\ \text{V}$.

When comparing oxidation reactions, the one with the less negative (higher) oxidation potential occurs more readily. Here, $-1.23\ \text{V}$ is greater than $-1.36\ \text{V}$, so you might think water oxidation to $O_2$ should happen first.

But there's a catch: the $O_2$ evolution reaction involves 4 electrons and has a significant overpotential on platinum electrodes. Overpotential is an extra voltage needed to overcome the kinetic barrier for a reaction to occur at a practical rate. On platinum, the overpotential for oxygen evolution is substantial (around 0.4–0.6 V), while for chlorine evolution it is very small. …

The key to this question lies in Faraday’s laws of electrolysis — specifically, the idea that the amount of charge needed to reduce or oxidise a substance is directly proportional to the number of electrons transferred per mole. But before we plug numbers, let’s understand why the electron count is what it is.

Why the electron count matters

Faraday’s first law says: the mass of a substance liberated at an electrode is proportional to the quantity of electricity passed. But for a mole of ions, the charge required is simply:

where $n$ is the number of electrons transferred per ion (or molecule) and $F$ is the Faraday constant (≈ 96485 C/mol). So the problem reduces to finding $n$ for the half‑reaction:

Step‑by‑step reasoning

1. Identify the oxidation states In $\text{MnO}_4^-$, oxygen is always –2 (except in peroxides, but not here). Let the oxidation state of Mn be $x$.

So Mn is in the +7 state.

In $\text{Mn}^{2+}$, the oxidation state is clearly +2.

2. Find the change in oxidation state The change is from +7 to +2:

A decrease of 5 in oxidation number means the Mn atom has gained 5 electrons.

3. Balance the half‑reaction (to confirm) …

Electrolysis involves using electrical energy to drive non-spontaneous chemical reactions. When an aqueous solution is electrolyzed, there's often a competition between water molecules and the dissolved ions to be oxidized or reduced at the electrodes. The outcome depends on their standard electrode potentials and kinetic factors like overpotential.

At the cathode (where reduction occurs), the species with a more positive (or less negative) standard reduction potential will be preferentially reduced.

At the anode (where oxidation occurs), the species with a more negative (or less positive) standard reduction potential (or more positive standard oxidation potential) will be preferentially oxidized.

Let's break down the process for aqueous NaCl.

1. Identify Species Present in Aqueous NaCl Solution:

   When NaCl is dissolved in water, it dissociates into $Na^+$ and $Cl^-$ ions. Water itself is also present. So, the species available for reaction are:

   • Cations: $Na^+(aq)$, $H^+(aq)$ (from water autoionization)
   • Anions: $Cl^-(aq)$, $OH^-(aq)$ (from water autoionization)
   • Solvent: $H_2O(l)$

2. Reactions at the Cathode (Reduction):

   The cathode is the negative electrode where reduction takes place. Possible species to be reduced are $Na^+$ ions and $H_2O$ molecules.

   • Reduction of $Na^+$: $Na^+(aq) + e^- \rightarrow Na(s)$ $E^\circ = -2.71 \text{ V}$
   • Reduction of $H_2O$: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ $E^\circ = -0.83 \text{ V}$ (at standard conditions, $1 \text{ M } OH^-$)

   Comparing the standard reduction potentials, $-0.83 \text{ V}$ is significantly less negative (more positive) than $-2.71 \text{ V}$. This means that water is much easier to reduce than $Na^+$ ions. Therefore, $H_2$ gas will be liberated at the cathode.

   Watch out

   A common misconception is that $Na^+$ will be reduced because it's a cation. However, in aqueous solutions, the reduction potential of water often dictates the cathode product if the metal ion is very reactive (has a very negative reduction potential).

3. Reactions at the Anode (Oxidation):

   The anode is the positive electrode where oxidation takes place. Possible species to be oxidized are $Cl^-$ ions and $H_2O$ molecules.

   • Oxidation of $Cl^-$: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ $E^\circ_{ox} = -1.36 \text{ V}$ (or $E^\circ_{red} = +1.36 \text{ V}$)
   • Oxidation of $H_2O$: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$ $E^\circ_{ox} = -1.23 \text{ V}$ (or $E^\circ_{red} = +1.23 \text{ V}$)

   Based purely on standard oxidation potentials, $H_2O$ (with $E^\circ_{ox} = -1.23 \text{ V}$) appears to be easier to oxidize than $Cl^-$ (with $E^\circ_{ox} = -1.36 \text{ V}$) because $-1.23 \text{ V}$ is less negative (more positive). This would suggest $O_2$ should be liberated.

   However, there's a crucial kinetic factor called overpotential.

   Important

   Overpotential: The extra voltage required beyond the theoretical standard electrode potential to initiate a reaction at a reasonable rate. This is particularly significant for the evolution of gases like $O_2$ and $H_2$ on certain electrode surfaces. …

Why this approach works

Electrolysis problems hinge on a simple chain: chemical change → electrons transferred → charge required. Faraday's laws of electrolysis tell us that the amount of substance transformed at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The bridge between chemistry and electricity is the mole of electrons, quantified by the Faraday constant $F = 96{,}485 \, \text{C/mol}$.

When we oxidize water to oxygen, we're stripping electrons away. The first step is always to write the balanced half-reaction so we know exactly how many electrons are involved per mole of product.

Step-by-step solution

1. Write the oxidation half-reaction for water.

Water molecules lose electrons to form oxygen gas. In acidic or neutral conditions, the half-reaction is:

This tells us that producing 1 mole of $O_2$ requires the removal of 4 moles of electrons from 2 moles of water.

2. Identify what the question asks.

The question asks for the charge needed to oxidize 1 mol of $H_2O$, not 1 mol of $O_2$. This is a crucial distinction.

From the half-reaction above, 2 mol of $H_2O$ produce 1 mol of $O_2$ and release 4 mol of electrons. Therefore, 1 mol of $H_2O$ releases:

3. Convert moles of electrons to charge using Faraday's constant.

The charge $Q$ required is:

$Q = n \cdot F$ …