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What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

Short Answer

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Electrochemistry is the study of the conversion between chemical energy and electrical energy, which often involves redox reactions - chemical reactions where electrons transfer between species. A galvanic cell, also known as a voltaic cell, is an electrochemical cell in which a spontaneous redox reaction produces electrical energy. On the other hand, an electrolytic cell uses external electrical energy to drive a non-spontaneous redox reaction. The primary difference lies in the direction of energy conversion: galvanic cells convert chemical energy into electrical energy, while electrolytic cells convert electrical energy into chemical energy.

Step by step solution

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1. Defining Electrochemistry

Electrochemistry is a branch of chemistry that deals with the relationship between electrical energy and chemical reactions. In other words, it is the study of the conversion between chemical energy and electrical energy. It plays a vital role in a wide variety of fields, including batteries, fuel cells, and even corrosion.
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2. Understanding Redox Reactions

Redox (short for reduction-oxidation) reactions are a type of chemical reaction that involve a transfer of electrons between two chemical species. In these reactions, one species undergoes oxidation (loses electrons) while the other undergoes reduction (gains electrons). This electron transfer can result in the generation of an electric current or the formation of new compounds. Examples of redox reactions include combustion and many cellular processes such as respiration and photosynthesis.
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3. Galvanic Cells

A galvanic cell, also known as a Voltaic cell, is a type of electrochemical cell in which a spontaneous redox reaction produces electrical energy. In a galvanic cell, the oxidation and reduction processes occur separately in two different compartments (called half-cells) that are connected by an external circuit and a salt bridge. The species undergoing oxidation loses electrons which travel through the external circuit to the reduced species, generating an electric current. An example of a galvanic cell is the simple zinc-copper cell in which the spontaneous zinc oxidation and copper reduction generate electrical energy that can be used to power an external device.
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4. Electrolytic Cells

An electrolytic cell, on the other hand, is a type of electrochemical cell where an electrical energy is used to drive a non-spontaneous redox reaction. In this case, the applied voltage causes the oxidation and reduction reactions to occur; they would not occur without the external energy input. Electrolytic cells are widely used in various applications such as the electroplating of metals and the production of various chemicals by electrolysis, like aluminum and chlorine.
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5. Comparing Galvanic and Electrolytic Cells

The main difference between galvanic and electrolytic cells lies in the direction of the energy conversion: - In a galvanic cell, a spontaneous redox reaction generates electrical energy, i.e., chemical energy is converted into electrical energy. - In an electrolytic cell, an external electrical energy source drives a non-spontaneous redox reaction, i.e., electrical energy is converted into chemical energy. Other key differences include: - Galvanic cells produce electricity, while electrolytic cells consume electricity. - In a galvanic cell, the anode (where oxidation occurs) is the negative electrode, and the cathode (where reduction occurs) is the positive electrode; In an electrolytic cell, the anode is the positive electrode, and the cathode is the negative electrode. - Galvanic cells require a spontaneous redox reaction, while electrolytic cells utilize a non-spontaneous redox reaction.

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Most popular questions from this chapter

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{kJ} .\) What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

A galvanic cell is based on the following half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{array}$$ In this cell, the silver compartment contains a silver electrode and excess \(\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. Assuming \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\space \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{r}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)\)

What volume of \(\mathrm{F}_{2}\) gas, at \(25^{\circ} \mathrm{C}\) and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for \(2.00 \mathrm{h} ?\) What mass of potassium metal is produced? At which electrode does each reaction occur?

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