Problem 13 - Chemistry in 30 days

Question

How many faradays are required to reduce \(\mathrm{Sn}^{4+}\) to \(\mathrm{Sn}^{2+} ?\) (a) \(2.0\) (b) \(4.0\) (c) \(1.0\) (d) \(6.0\)

Step-by-Step Solution

Answer

2.0 faradays are required.

Step 1: Determine the Charge Transfer

To find how many faradays are required to reduce \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \), we first need to determine the overall charge transfer in the process. The reduction from \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \) involves gaining 2 electrons since: \[ \mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+} \].

Step 2: Define a Faraday

A faraday is a unit of electric charge equal to one mole of electrons, or approximately 96,485 coulombs. It represents the charge of one mole of electrons transferred in electrochemical reactions.

Step 3: Calculate Faradays Required

Since the reduction of \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \) requires 2 electrons, it follows that 1 mole of \( \mathrm{Sn}^{4+} \) requires 2 faradays (because each faraday corresponds to 1 mole of electrons).

Key Concepts

Electrochemical Reactions

Electrochemical reactions are processes where chemical reactions create electricity, or electricity results in chemical changes. This forms the basis of batteries and electrolysis. These reactions occur in a unique setup commonly known as an electrochemical cell, which consists of two electrodes and an electrolyte. In these cells, oxidation (loss of electrons) and reduction (gain of electrons) reactions happen at the electrodes. For instance, when you consider reducing \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \), this reduction reaction involves adding electrons to \( \mathrm{Sn}^{4+} \). The process is spontaneous in a battery and requires energy input in electrolysis.

• Oxidation always occurs at the anode.
• Reduction always takes place at the cathode.

Understanding these reactions is crucial because they explain how batteries work and how electroplating can change materials. Electrochemical reactions convert electrical energy into chemical energy and vice versa.

Charge Transfer

The concept of charge transfer is important in understanding electrochemical reactions. It refers to the movement of electrons from one atom, ion, or molecule to another. In the context of reducing \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \), charge transfer involves moving two electrons from an electron donor to \( \mathrm{Sn}^{4+} \), allowing it to become \( \mathrm{Sn}^{2+} \). This electron movement is essential because it changes the oxidation state of the ion.
Charge transfer is calculated in moles, which relate directly to the number of electrons moving in an electrochemical cell.

• 1 electron moved equals 1 elementary charge.
• 1 mole of electrons moved equals 1 faraday of charge.

Each movement alters the properties and behaviors of the involved substances, driving the foundational principles of electrochemistry.

Moles of Electrons

The mole is a fundamental concept in chemistry, representing a specific number of entities, typically atoms or molecules. When dealing with electrochemical reactions, the flow of moles of electrons is crucial.
The reduction of \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \) involves adding 2 electrons, which equates to 2 moles of electrons for each mole of \( \mathrm{Sn}^{4+} \). Here's why moles of electrons are critical:

• They quantify the charge transfer in electrochemical reactions.
• 1 mole of electrons corresponds to 1 faraday, aligning physical charge with chemical amounts.

In practical terms, calculating the moles of electrons tells you how much electric charge is required to cause a specific chemical change, as in the example where 2 moles of electrons or 2 faradays are needed for the reduction of \( \mathrm{Sn}^{4+} \) to \( \mathrm{Sn}^{2+} \). This conversion is crucial for applications in designing batteries and industrial electrochemical processes.