Question

Use Table 17.1 to answer the following questions. Use LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) For the half reaction $$ \frac{1}{2} \mathrm{Br}_{2}(l)+e^{-} \longrightarrow \mathrm{Br}^{-}(a q) $$ \(E_{\text {red }}^{\circ}\) ______ \(1.077 \mathrm{~V}\) (b) For the reaction $$ 2 \mathrm{Br}^{-}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \operatorname{Br}_{2}(l)+\mathrm{Co}(s) $$ \(E^{\circ}\) _______ 0. (c) If the half reaction $$ \mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) $$ is designated as the new standard where \(E_{\text {red }}^{\circ}\) is 0.00 , then \(E_{\text {red }}^{\circ}\) for \(2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)\) is ______ 0. (d) For the reaction $$ 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Co}(s) \longrightarrow 2 \mathrm{Cr}(s)+3 \mathrm{Co}^{2+}(a q) $$ the number of electrons exchanged is _______ 6. (e) For the reaction described in (d), the number of coulombs that passes through the cell is ______ \(9.648 \times 10^{4}\).

Short answer

#Answer# #tag_title# Part (a) - Comparing the standard reduction potential#tag_content# Using the standard potential values from Table 17.1, we can say that (arbitrary symbol) 1.077 V. #tag_title# Part (b) - Determine standard cell potential#tag_content# Using the values from Table 17.1, the calculated standard cell potential (E°) is (calculated value) V. #tag_title# Part (c) - Determine the new standard reduction potential#tag_content# The new standard reduction potential for the given reaction is (new value) V. #tag_title# Part (d) - Determine the number of electrons exchanged#tag_content# The number of electrons exchanged in the given reaction is (number of electrons) e^-. #tag_title# Part (e) - Determine the number of coulombs passed through the cell#tag_content# The total number of coulombs passing through the cell is (total coulombs) C.

Step-by-step solution

Part (a) - Comparing the standard reduction potential

For the given half reaction, we need to compare the \(E_{red}^\circ\) value to 1.077 V. We will use Table 17.1 for this, use any symbol stated above to complete the statement.

Part (b) - Determine standard cell potential

We are given the overall reaction: $$ 2 Br^-(aq) + Co^{2+}(aq) \longrightarrow Br_{2}(l) + Co(s) $$ We will split this reaction into two half reactions and use their standard reduction potentials in Table 17.1 to find the overall standard cell potential (E°). The half reactions are: 1) $$ \frac{1}{2}Br_{2}(l) + e^- \longrightarrow Br^-(aq) $$ 2) $$ Co^{2+}(aq) + 2e^- \longrightarrow Co(s) $$ Calculate \(E^{\circ}\): $$ E^{\circ} = E^{\circ}_{\text{Red(2)}} - E^{\circ}_{\text{Red(1)}} $$

Part (c) - Determine the new standard reduction potential

The given standard reduction potential for the new reaction: $$ Co^{2+}(aq)+2e^- \longrightarrow Co(s) $$ is designated as zero. We need to find the standard reduction potential for the reaction: $$ 2H^{+}(aq) + 2e^- \longrightarrow H_{2}(g) $$ Use Table 17.1 to find the standard reduction potential of the given reaction and report the new value.

Part (d) - Determine the number of electrons exchanged

For the reaction: $$ 2Cr^{3+}(aq) + 3Co(s) \longrightarrow 2Cr(s) + 3Co^{2+}(aq) $$ We need to determine the number of electrons exchanged in this reaction. Break it down into half reactions to better understand the electron transfer process. The half reactions are: 1) $$ Cr^{3+}(aq) + 3e^- \longrightarrow Cr(s) $$ 2) $$ Co^{2+}(aq) + 2e^- \longrightarrow Co(s) $$ Calculate the number of electrons exchanged by comparing the half reactions and report that value.

Part (e) - Determine the number of coulombs passed through the cell

In the given reaction in part (d), the number of electrons exchanged is determined in the previous part. To calculate the number of coulombs, we use the Faraday's constant (F) which is equal to the charge of a mole of electrons: $$ F = 9.648 \times 10^4 C/mol $$ Number of coulombs passed through the cell can be calculated using the formula: $$ Q = n \times F $$ where n is the number of moles of electrons exchanged and Q is the total charge passing through the cell in coulombs. Use the value of electrons exchanged from part (d) to calculate the total coulombs passed through the cell.

Key concepts

Understanding Electrochemistry

Electrochemistry is a branch of chemistry that deals with the relationship between electrical potential, or voltage, and chemical change. It involves the study of chemical reactions that involve the transfer of electrons, and these processes are crucial to numerous modern technologies like batteries, fuel cells, and corrosion prevention.

In electrochemistry, reactions are categorized as either oxidation, where an atom or molecule loses electrons, or reduction, where an atom or molecule gains electrons. Importantly, these two processes occur simultaneously in what is called a redox reaction. The electrons transferred can be measured in volts, using a reference such as the standard hydrogen electrode, which gives us the standard reduction potentials (E_red°) for various half-reactions.

Galvanic Cells Explained

A Galvanic cell, also known as a voltaic cell, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It usually consists of two different metals connected by a salt bridge, or individual half-cells linked by a conductive solution.

Each metal in a Galvanic cell has its own standard reduction potential. By comparing the potentials of two half-reactions, one can determine which metal acts as an anode (oxidation occurs) and which acts as a cathode (reduction occurs). The difference in potential between anode and cathode is equal to the electrical voltage that the cell can produce.

Key Components of a Galvanic Cell:

• Anode: Where oxidation occurs, and electrons are produced.
• Cathode: Where reduction occurs, and electrons are consumed.
• Electrolyte: Conductive solution allowing ion transfer.
• Salt bridge: Maintains electrical neutrality.

By connecting the anode and cathode externally, electrons can flow and produce useful electrical work.

Electron Exchange in Reactions

Electron exchange is the fundamental process that drives electrochemical reactions. It is the movement of electrons from one reactant (oxidizing agent) to another (reducing agent), leading to changes in the oxidation states of the atoms or ions involved.

The number of electrons exchanged can be determined by balancing the oxidation and reduction half-reactions for mass and charge. This process ensures that the net charge before and after the reaction remains the same, which is essential for the conservation of charge. In an electrochemical cell, this electron exchange generates an electrical current which can be measured and harnessed for power.

Faraday's Law of Electrolysis

Faraday's law of electrolysis quantitatively relates the amount of electric charge passed through a solution to the amount of substance that is deposited at the electrode. Michael Faraday formulated these laws, which are essential in the field of electrochemistry, particularly in electrolysis.

The first law states that the mass of a substance altered at an electrode during electrolysis is proportional to the quantity of charge passed. The second law states that for the same quantity of charge passed, the mass of substance altered at the electrode is proportional to its equivalent weight.

mathematically, the law can be expressed as:
$$ Q = n \times F $$where Q is the total charge in coulombs, n is the number of moles of electrons, and F is the Faraday constant, approximately equal to 96485 coulombs per mole of electrons. This principle is crucial in calculating the amount of reactant consumed or product generated in an electrochemical cell, and plays a role in battery design, metal plating processes, and the production of pure substances.