I Suppose \(E_{\text {red }}^{\circ}\) for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) were taken to be \(0.300 \mathrm{~V}\) instead of \(0.000 \mathrm{~V}\). What would be (a) \(E_{\text {ox }}^{o}\) for \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}\) ? (b) \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-} ?\) (c) \(E^{\circ}\) for the cell in \(24(\mathrm{c})\) ? Compare your answer with that obtained in \(24(\mathrm{c})\).

Short Answer

Expert verified
Answer: The new standard cell potential is 1.080 V.

Step by step solution

01

Calculate the standard oxidation potential for hydrogen reaction given new std. reduction potential for hydrogen electrode

We are given a new standard reduction potential (\(E_{\text {red }}^{\circ}\)) for the hydrogen electrode and asked to find the corresponding standard oxidation potential (\(E_{\text {ox }}^{\circ}\)). We can simply find it by taking the negative of the given value, as the standard oxidation and reduction potentials are related by: \(E_{\text {ox }}^{\circ} = - E_{\text {red }}^{\circ}\) Calculate the standard oxidation potential for \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}\) reaction: \(E_{\text {ox }}^{\circ} = - 0.300 \mathrm{~V}\) = \(-0.300 \mathrm{~V}\)
02

Calculate the standard reduction potential for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}\) reaction

We have already calculated the new standard oxidation potential for the hydrogen electrode as \(-0.300 \mathrm{~V}\). Now, to find the new standard reduction potential for the bromine and bromide ions redox reaction, we need to find their half-reaction potential. Given this redox reaction: \(\mathrm{Br}_{2} + 2 \mathrm{e^{-}} \longrightarrow 2 \mathrm{Br}^{-}\) with \(E^{\circ} = 1.080 \mathrm{~V} (from\ merck\ index, 13th\ edition)\) Add the new standard oxidation potential of the hydrogen electrode to the bromine standard reduction potential. New \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}\) = \(1.080 - 0.300 = 0.780 \mathrm{~V}\)
03

Calculate the new cell potential

Now, we have the new \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}\) reaction. We are asked to find the cell potential for the cell in 24(c). The previous cell potential in 24(c) was given as: \(E_{cell}^{\circ} = E_{\text {red }}^{\circ} - E_{\text {ox }}^{\circ} = 1.080 \mathrm{~V}\) (the standard hydrogen potential was \(0.000 \mathrm{~V}\)) Now, we calculate the new cell potential using the same formula: \(E^{\circ}_{new \ cell} = E_{\text {red }}^{\circ} - E_{\text {ox }}^{\circ} = 0.780 \mathrm{~V} - (-0.300 \mathrm{~V})\) \(E^{\circ}_{new \ cell} = 1.080 \mathrm{~V}\) Comparing it with the initial standard cell potential, we see that there is no change, so the new cell potential is the same as the previous one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrochemistry
Electrochemistry is a fascinating field of science that deals with the relationships between electricity and chemical reactions. At the very heart of electrochemistry are the processes of oxidation and reduction, collectively known as redox reactions. When these reactions take place, electrons are transferred between substances, leading to the creation of electrical energy or the use of electrical energy to drive chemical changes. This interplay is essential in a wide array of technologies, from batteries and fuel cells to electrolysis and corrosion prevention.

Understanding the standard reduction potential, denoted as \(E_{red}^{\circ}\), is crucial since it allows us to predict the direction of electron flow in a redox reaction. This value, often provided at standard conditions of 298 K, 1 atm, and 1 M concentration, is a measure of the tendency of a species to be reduced. In practice, it means how likely it is to gain electrons. If we alter the standard reduction potential for a half-reaction, as was done hypothetically for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) in the exercise, it can significantly impact the electrochemical system's behavior.
Half-Reactions
Half-reactions are the decomposition of redox reactions into two parts: one that involves the loss of electrons, called oxidation, and one that involves the gain of electrons, called reduction. Each half-reaction has an associated potential, known as the standard reduction potential \(E_{red}^{\circ}\) for the reduction half, and the standard oxidation potential \(E_{ox}^{\circ}\) for the oxidation half.

In the exercise, we examined the implications of changing the standard reduction potential for hydrogen. By convention, the standard reduction potential for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) is set as 0.000 V as a reference. However, if this value is hypothetically altered to 0.300 V, it inversely changes the standard oxidation potential—the potential for the reverse reaction, \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}\), is consequently \(-0.300 V\). It demonstrates the fundamental principle that oxidation and reduction potentials for a given half-reaction are equal in magnitude but opposite in sign.
Cell Potential
Cell potential, also recognized as the electromotive force (EMF) of a cell, denotes the driving force behind the movement of electrons from one electrode to another in an electrochemical cell. Cell potential is the difference between the reduction potentials of the cathode and the anode.

The formula to calculate cell potential is given by:
\[E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\]

In our hypothetical exercise scenario, the potential for the hydrogen reaction is changed, but curiously, the overall cell potential remains unchanged at 1.080 V when the new standard reduction and oxidation potentials are applied to the bromine reaction. This situation illustrates a key concept: the cell potential depends on the relative potentials of both half-reactions, and altering one half-reaction's potential does not necessarily change the overall cell potential if the relative difference remains the same.
Redox Reactions
Redox reactions are a type of chemical reaction in which oxidation and reduction occur simultaneously. These reactions are fundamental to numerous natural and industrial processes, including energy production in living organisms and the functioning of batteries.

The substance that donates electrons is being oxidized and is referred to as the reducing agent. Conversely, the substance that receives electrons is being reduced and acts as the oxidizing agent. Electrochemistry quantifies these reactions through potentials, with more positive reduction potentials indicating a stronger tendency for reduction to occur. In the bromine reaction mentioned in the exercise, \(\mathrm{Br}_{2} + 2e^{-} \longrightarrow 2\mathrm{Br}^{-}\), the standard reduction potential is adjusted based on the altered hydrogen potential. Understanding how changes in standard reduction potential influence redox reactions and cell potential is key for students studying electrochemistry and for any practical applications involving electrochemical cells.

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Most popular questions from this chapter

Which of the changes below will increase the voltage of the following cell? $$ \text { Co }\left|\mathrm{Co}^{2+}(0.010 M) \| \mathrm{H}^{+}(0.010 \mathrm{M})\right| \mathrm{H}_{2}(0.500 \mathrm{~atm}) \mid \mathrm{Pt} $$ (a) Increase the volume of \(\mathrm{CoCl}_{2}\) solution from \(100 \mathrm{~mL}\) to \(300 \mathrm{~mL}\). (b) Increase \(\left[\mathrm{H}^{+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\) (c) Increase the pressure of \(\mathrm{H}_{2}\) from \(0.500 \mathrm{~atm}\) to \(1 \mathrm{~atm}\). (d) Increase the mass of the Co electrode from \(15 \mathrm{~g}\) to \(25 \mathrm{~g}\). (e) Increase \(\left[\mathrm{Co}^{2+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\).

In biological systems, acetate ion is converted to ethyl alcohol in a two- step process: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}(a q)+\mathrm{H}_{2} \mathrm{O}\) \(E^{o \prime}=-0.581 \mathrm{~V}\) \(\mathrm{CH}_{3} \mathrm{CHO}(a q)+2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \quad E^{o \prime}=-0.197 \mathrm{~V}\) \(\left(E^{\circ \prime}\right.\) is the standard reduction voltage at \(25^{\circ} \mathrm{C}\) and \(\mathrm{pH}\) of \(7.00 .\) ) (a) Calculate \(\Delta G^{\circ \prime}\) for each step and for the overall conversion. (b) Calculate \(E^{\circ \prime}\) for the overall conversion.

A baby's spoon with an area of \(6.25 \mathrm{~cm}^{2}\) is plated with silver from \(\mathrm{AgNO}_{3}\) using a current of \(2.00 \mathrm{~A}\) for two hours and 25 minutes. (a) If the current efficiency is \(82.0 \%\), how many grams of silver are plated? (b) What is the thickness of the silver plate formed \(\left(d=10.5 \mathrm{~g} / \mathrm{cm}^{3}\right)\) ?

Which of the following reactions is (are) spontaneous at standard conditions? (a) \(\mathrm{Zn}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Fe}^{2+}(a q)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{H}_{2}(g)\) (c) \(2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Br}_{2}(l)+2 \mathrm{I}^{-}(a q)\)

Consider the reaction $$ \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Ag}(s)+2 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{AgBr}(s)+\mathrm{H}_{2} \mathrm{~S}(a q) $$ At what \(\mathrm{pH}\) is the voltage zero if all other species are at standard concentrations?

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