Welcome to our study guide on electrochemical cells, including galvanic and electrolytic cells! This guide will help you navigate through the concepts of standard electrode potentials, use the Nernst equation for non-standard conditions, and appreciate both the differences and similarities between galvanic and electrolytic cells. So let's jump into this electric topic!

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🌐 Principles of Galvanic and Electrolytic Cells

Electrochemical cells are like tiny powerhouses where redox reactions occur. There are two main types of electrochemical cells: galvanic cells and electrolytic cells.

🎛 Galvanic (Voltaic) Cells

Galvanic cells are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions. These cells are fundamental components of batteries! Here's a breakdown of each component and its role in the process:

1. Anode - the negative electrode in a Galvanic cell. It is where oxidation occurs, meaning it is the site of electron loss. The material that makes up the anode is the reductant in the cell's chemical reaction. Electrons flow from the anode through the external circuit to the cathode.
2. Cathode - the positive electrode. It is where reduction happens, which means it is the site where electrons are gained. The cathodic material is the oxidant, accepting electrons that have traveled through the external circuit from the anode.
3. Electrolyte - a conductive solution that contains ions. The electrolyte allows ions to move between the anode and cathode compartments, which is necessary to maintain charge neutrality in the cell as electrons flow through the external circuit.
4. Electrical circuit - a conductive path that connects the anode and cathode, allowing electrons to flow from the anode to the cathode. This flow of electrons is what we use as electrical energy.
5. Salt bridge - serves to balance the ion distribution by allowing ions to move between the anode and cathode compartments. This prevents the buildup of positive or negative charges in either compartment, which would stop the flow of electrons and, consequently, the electrical current

Image courtesy of Chemistry LibreTexts.

🔋 Electrolytic Cells

Electrolytic cells are a type of electrochemical cell used for electrolysis, where electrical energy is converted into chemical energy by driving non-spontaneous reactions. Unlike galvanic cells, which generate electrical energy from spontaneous chemical reactions, electrolytic cells require an external power source to force a chemical change.

💭 Comparison: Galvanic vs Electrolytic Cells

Here's what sets these two apart:

• Flow Direction: Electrons flow from anode to cathode in both types; however...
  • In galvanic cells, it's due to natural redox reactions.
  • In electrolytic cells, it's forced by an external voltage source.

Practical Applications:

• Galvanic Cells: Used in batteries powering devices from watches to cars!
• Electrolytic Cells: Critical in industries for electroplating metals or producing substances like chlorine or aluminum through electrolysis.

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⚡ Standard Electrode Potentials (E°)

Standard electrode potentials (E° values) measure the tendency of a chemical species to be reduced under standard conditions (1 M concentration for solutions, 1 atm pressure for gases, at 25°C).

🔍 The electrochemical series lists elements according to their standard reduction potentials. Metals at the top have stronger tendencies to be oxidized than those below them.

Calculating Cell Potential

For a galvanic cell: E° cell = E° cathode - E° anode

✅ If the E° cell is positive, the reaction is spontaneous!

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🧪 The Nernst Equation

When conditions aren't standard (different concentrations, pressures, temperatures), we use the Nernst equation:

Where:

• E = reduction potential under non-standard conditions
• E° = standard cell potential
• R = universal gas constant (8.314 Jmol^-1K^-1)
• T = temperature in Kelvin
• n = ion charge (number of moles of electrons)
• F = Faraday's constant (96,485 C/mol)
• Q = reaction quotient

Use this equation to calculate how changing concentrations or temperatures can affect cell voltage.

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🔬 Experimental Setup & Observations

Setting up your simple electrochemical experiments in class can be enlightening! You might observe:

• Metal deposition on electrodes during electrolysis.
• Gas evolution at electrodes (like hydrogen or oxygen).

Be sure to measure voltage and current accurately—it will test your understanding of theoretical predictions versus real-life results!

Guided Practice Questions:

1. Calculate the standard cell potential for a galvanic cell composed of zinc $\text{Zn}/\text{Zn}^{2+}$ with $\text{E}^{\circ}_{\text{anode}}=-0.76\text{ V}$ as an anode and copper $\text{Cu}/\text{Cu}^{2+}$ with $\text{E}^{\circ}_{\text{cathode}}=+0.34\text{ V}$ as a cathode. Is this reaction spontaneous?

   Answer:

   Using the formula $\text{E}^{\circ}_{\text{cell}}=\text{E}^{\circ}_{\text{cathode}}-\text{E}^{\circ}_{\text{anode}}$, we should be able to plug in the given numbers. Your equation should look like this: $\text{E}^{\circ}_{\text{cell}} = 0.34\text{ V} - (-0.76\text{ V})=0.34\text{ V} +0.76\text{ V}=\boxed{+1.10\text{ V}}$. Because the $\text{E}^{\circ}_{\text{cell}}$ is positive, the reaction is spontaneous.

2. For an electrolytic process using a silver cation solution $\text{Ag}^+$, if we apply a voltage greater than its reduction potential measured at $+0.80\text{ V}$ under standard conditions, will the silver metal plate out onto the cathode? Explain your reasoning based on what drives electrolysis.

   Answer:

   Yep! If a voltage greater than $+0.80\text{ V}$ is applied, the silver metal will out onto the cathode. This is because the greater voltage, which is still over $+0.80\text{ V}$, will help a non-spontaneous reaction occur and help the $\text{Ag}^+$ ions become regular $\text{Ag}$, which deposits on the silver plate on the cathode.

3. Calculate the cell potential for a voltaic cell based on the following half-reaction and conditions: $\text{Fe}^{3+}_{(aq)}+e^- \rightarrow \text{Fe}^{2+}_{(aq)}$ Given: $\text{E}^{\circ}_{\text{cell}}=+0.77\text{V}$ (standard cell potential)

   $[\text{Fe}^{3+}]=1.0\text{M}$ (standard state) $[\text{Fe}^{2+}]=0.1\text{M}$ (non-standard state) Temperature ($T$) $=298\text{ K}$

   Answer: First, we need to determine the reaction quotient ($Q$) using the concentrations of $\text{Fe}^{3+}$ and $\text{Fe}^{2+}$ ions:

   $Q=[\text{Fe}^{3+}]\cdot[\text{Fe}^{2+}]=1.0\cdot0.1$

   Next, we'll apply the Nernst equation:

   $\text{E}_{\text{cell}}=\text{E}^{\circ}_{\text{cell}}-\frac{0.0592\text{ V}}{n}\cdot\text{log}(Q)$

   Since the half-reaction involves the transfer of one mole of electrons ($n=1$), we can plug in the values: $\text{E}_{\text{cell}}=0.77\text{ V}−0.0592\cdot \text{log}(0.1)$ Calculate the natural logarithm of $Q$:

   $\text{log}(0.1)=-1$

   Substitute into the equation:

   $\text{E}_{\text{cell}}=0.77\text{ V}−0.0592\text{ V}\cdot (-1)=0.77\text{ V}+0.0592\text{ V}=\boxed{0.83\text{ V}}$

Remember chemistry enthusiasts; understanding comes from practice! Dive into these questions before advancing further into applications such as corrosion prevention or fuel cell technologies!

Happy Studying! ⚗️🔋