Redox Reactions

Based on the observations of the biologist Galvani, the Italian physicist Volta conducted research that led him to establish his "series of metal tensions" in 1793.
In 1800, he discovered the electric battery that bears his name. These discoveries laid the foundation for the interpretation of redox phenomena.

I. First Concept of Redox Reactions

1. Can a metal react with a metallic ion?

zinc and copper oxidation experiment at t = 0s

Observations for experiment 1:

The blue coloration of the solution is significantly reduced
A red-orange deposit, characteristic of metallic copper (Cu), is observed
A white precipitate, characteristic of Zn²⁺ ions, is formed

Observations for experiment 2:

No change, indicating that no chemical transformation occurred

Reminder: The blue color is characteristic of Cu²⁺ ions in water.

Now, it's your turn to interpret experiment 1 by selecting the 10 correct answers below!

Correct answer(s): 0/10Errors: 0

Congratulations! You found all the correct answers!

2. Writing the Reaction Equation

We proceed step by step:

We write two half-equations that correspond to the loss or gain of one or more electrons.
We sum these two half-equations to obtain the overall reaction equation.

Cu²⁺ + 2e^- → Cu (Reduction: Cu²⁺ gains 2 electrons)
Zn → Zn²⁺ + 2e^- (Oxidation: Zn loses 2 electrons)

Cu²⁺ + Zn + 2e^-❌→ Cu + Zn²⁺ + 2e^-❌ (Overall redox equation)
Cu²⁺ + Zn → Cu + Zn²⁺ (Overall redox equation)

Note: The electrons cancel out directly. If this is not the case,
one of the two half-equations must be multiplied by a coefficient to balance the electrons.

In experiment 1, the electrons lost by Zn are captured by Cu²⁺.
The oxidant Cu²⁺ is transformed into Cu, which is a reductant.
Thus, we have the redox pair Cu²⁺ / Cu.
Similarly, we have the redox pair Zn²⁺ / Zn.
A redox pair is always written in the form oxidant / reductant.

II. In General

1. Basic Concepts

A chemical species capable of donating one or more electrons is a reductant.
A chemical species capable of accepting one or more electrons is an oxidant.
During a redox reaction, the oxidant of one pair is reduced, and the reductant of another pair is oxidized.

2. A Rule: Can We Predict the Spontaneous Reaction Between Two Redox Pairs?

Experiment 2 shows that simply combining an oxidant and a reductant does not necessarily lead to a chemical reaction.

Volta's work led to the association of a standard redox potential E⁰ (in Volts (V)) with each oxidant/reductant pair.
Using these values, it was possible to rank all the pairs.

The higher the E⁰ value, the stronger the oxidant in the pair and the weaker the reductant.
Cu²⁺ is therefore a stronger oxidant than Zn²⁺.
Zn is a stronger reductant than Cu.
The spontaneous redox reaction between two pairs always occurs between the strongest reductant and the strongest oxidant
(this is known as the "gamma rule").
This explains why Cu²⁺ and Zn react, whereas Zn²⁺ and Cu do not.

3. Synthesis Exercise

a. Select the 3 metals that oxidize the least easily

Congratulations, you found it!
The highest potential corresponds to the weakest reductant, meaning the one that reacts the least.

b. Indicate whether a reaction is possible between:

Fe and Ag⁺?

Ag and Cu²⁺?

Cu and H⁺?

Fe and Zn²⁺?

Zn and H⁺?

c. Drag and drop to obtain the reaction equation between Ag⁺ and Fe

Congratulations, you found the correct solution

🔽 drag & drop 🔽

( → + e-)x

( + e- → )x

Fe + 2Ag⁺ → Fe²⁺ + 2Ag

III. Other Important Oxidant/Reductant Pairs

There are other redox reactions where metals do not take part. Acidified aqueous solutions are often used.

H₂O is present in the solvent
If acid is added, H⁺ is also present as it is responsible for the acidic nature of the solution
H₂O and H⁺ can therefore take part in the reaction equations

a. Drag and drop to obtain the possible reaction between the pairs I₂ / I^- and S₂O₈^2- / SO₄^-

Well done, you found the correct solution

🔽 drag / drop 🔽

( + e- → )x

( → + e-)x

S₂O₈^2- + 2I^- → 2SO₄^- + I₂

b. Drag and drop to obtain the possible reaction between the pairs MnO₄^- / Mn²⁺ and O₂ / H₂O₂

Well done, you found the correct solution

🔽 glisser / déposser 🔽

( + e- + → + )x

( → + + e-)x

10e^-❌ + 2MnO₄^- + 16H⁺❌6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂ + 10H⁺❌ +10e^-❌

2MnO₄^- + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂