/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 Predict whether the following re... [FREE SOLUTION] | 91影视

91影视

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Cu}\) to \(\mathrm{Cu}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), \((\mathbf{b})\) reduction of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}\) by \(\mathrm{H}_{2}\) (to form \(\mathrm{H}^{+}\) ), \(\left(\mathbf{c}\right.\) ) reduction of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) by \(\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{d})\) reduction of \(\mathrm{Ni}^{2+}\) to \(\mathrm{Ni}\) by \(\mathrm{Sn}^{2+}\left(\right.\) to form \(\left.\mathrm{Sn}^{4+}\right)\).

Short Answer

Expert verified
Among the four reactions, only the oxidation of Cu to Cu虏鈦 by I鈧 to form I鈦 (a) is spontaneous in acidic solution under standard conditions. E潞_cell for this reaction is positive (0.20 V). The other reactions (b, c, and d) have negative E潞_cell values, indicating that they are not spontaneous under the given conditions.

Step by step solution

01

Recall the standard potentials for the reactions under consideration

: To predict whether a given reaction will be spontaneous in acidic solution under standard conditions, we can refer to standard reduction potential tables. (a) Cu to Cu虏鈦 and I鈧 to I鈦. E潞(Cu虏鈦/Cu) = +0.34 V E潞(I鈧/I鈦) = +0.54 V (b) Fe虏鈦 to Fe and H鈧 to H鈦. E潞(Fe虏鈦/Fe) = -0.44 V E潞(H鈧/H鈦) = 0 V (since hydrogen is the reference) (c) I鈧 to I鈦 and H鈧侽鈧 to H鈧侽. E潞(I鈧/I鈦) = +0.54 V E潞(H鈧侽鈧/H鈧侽) = +1.78 V (d) Ni虏鈦 to Ni and Sn虏鈦 to Sn鈦粹伜. E潞(Ni虏鈦/Ni)= -0.26 V E潞(Sn鈦粹伜/Sn虏鈦) = +0.15 V
02

Calculate the overall cell potential for each reaction

: (a) E潞_cell = E潞(I鈧/I鈦) - E潞(Cu虏鈦/Cu) = 0.54 V - 0.34 V = 0.20 V Since E潞_cell is positive, the reaction is spontaneous. (b) E潞_cell = E潞(Fe虏鈦/Fe) - E潞(H鈧/H鈦) = -0.44 V - 0 V = -0.44 V Since E潞_cell is negative, the reaction is not spontaneous. (c) E潞_cell = E潞(I鈧/I鈦) - E潞(H鈧侽鈧/H鈧侽) = 0.54 V - 1.78 V = -1.24 V Since E潞_cell is negative, the reaction is not spontaneous. (d) E潞_cell = E潞(Ni虏鈦/Ni) - E潞(Sn鈦粹伜/Sn虏鈦) = -0.26 V - 0.15 V = -0.41 V Since E潞_cell is negative, the reaction is not spontaneous. In conclusion, among the four reactions, only the first reaction (oxidation of Cu to Cu虏鈦 by I鈧 to form I鈦) is spontaneous in acidic solution under standard conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Standard Reduction Potential
Standard reduction potentials ( E掳 ) are essential in determining the likelihood of a chemical reaction occurring under standard conditions (25掳C, 1 atm, and 1 M concentrations). A standard reduction potential is measured in volts and indicates a substance's ability to gain electrons, essentially how eagerly it wants to be reduced.

In electrochemistry, every substance has a specific reduction potential, which can be found in standard reduction potential tables. These tables list various chemical reactions and their respective E掳 values.

Here is how you can utilize them:
  • A higher E掳 value means a greater tendency to gain electrons and be reduced.
  • If comparing two half-reactions, the one with the higher E掳 value will act as the cathode (reduction site) in a galvanic cell.
By using standard reduction potentials, you can calculate the overall cell potential ( E掳_cell ) of a redox reaction, which determines if the reaction is spontaneous, as positive overall cell potentials indicate spontaneous reactions.
Spontaneous Reaction
A spontaneous reaction is one that occurs without needing additional energy once started. In electrochemistry, whether a reaction is spontaneous is determined using standard reduction potentials.

When you calculate the overall cell potential ( E掳_cell ) of a reaction, look for:
  • A positive E掳_cell indicates a spontaneous reaction.
  • A negative E掳_cell means the reaction is non-spontaneous, and an additional energy input is needed to start it.
To find E掳_cell , subtract the E掳 of the oxidation reaction from the E掳 of the reduction reaction. Redox reactions are the heart of electrochemical cells, and understanding whether they are spontaneous is crucial for predicting the behavior of these cells in real-world applications.
Oxidation
Oxidation is a chemical process where a substance loses electrons. In redox reactions, it occurs alongside reduction, which is the gain of electrons by another substance.

Remember:
  • The substance that loses electrons is oxidized and termed the reducing agent or reductant.
  • Every oxidation must be paired with a reduction, as electrons cannot exist free in solutions.
In an oxidation process, the oxidized substance's oxidation state increases.

For example, in the reaction highlighted in the solution, copper ( Cu ) is oxidized to copper ions ( Cu^{2+} ). This is essential to forming a complete redox reaction, where Iodine goes to I鈦 , paired with the reduction half to complete the loop of electron transfer. Understanding these electron transfers helps predict the pathway and feasibility of chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

The standard reduction potentials of the following half-reactions are given in Appendix E: $$ \begin{array}{l} \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) \\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Cd}(s) \\\ \mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{Sn}(s) \\\ \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s) \end{array} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell potential and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value.

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (b) \(2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{-5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ}\).

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.