Consider the table of the standard electrode potentials at 25 °C:Reduction half-reactionE° (V)Au13+ (aq) + 3e¯ → Au(s)1.50I03 (aq) + 6H*(aq) + 5e¯ → I2(aq) + 3H2O(1)1.202H+(aq) + 2e → H2(g)Cr³+ (aq) + 3e → Cr(s)-0.73 Identify whether Au or Cr or both or neither will dissolve in 1 M HIO3 .Both Au and Cr will dissolve.Only Cr will dissolve.Only Au will dissolve.Neither Au nor Cr will dissolve.SubmitPrevious AnswersCorrectHIO3 oxidizes metals through the following reduction half-reaction:I03 (aq) + 6H*(aq) + 5e →I2(aq) + 3H2O(1)Because this half-reaction is above the reduction of H+ in the table, HIO3 can oxidizesome metals (for example, copper or silver) that cannot be oxidized by HCl. In general,the metals with reduction half-reactions listed below the reduction of IO3- to I2 in thetable dissolve in HIO3 , whereas the metals listed above it do not. Therefore, Au(which has a reduction potential of 1.50 V) will not be oxidized, but Cr (which has areduction potential of –0.73 V) will be oxidized by both HCl and HIO3 .Part BEnter a balanced redox equation for the reaction that occurs.Express your answer as a balanced net ionic equation including phases.ΑΣφ

From given picture I am solving part B however you have not mentioned but it seems that part A is…

Consider the table of the standard electrode potentials at 25 °C: Reduction half-reaction E° (V) Au³+(aq) + 3e¯ → Au(s) 1.50 I03 (aq) + 6H*(aq) + 5e¯ →I2(aq) + 3H2O(1) 1.20 2H+(aq) + 2e → H2(g) Cr³+(aq) + 3e¯ → Cr(s) -0.73

Consider the table of the standard electrode potentials at 25 °C: Reduction half-reaction E° (V) Au³+(aq) + 3e¯ → Au(s) 1.50 I03 (aq) + 6H*(aq) + 5e¯ →I2(aq) + 3H2O(1) 1.20 2H+(aq) + 2e → H2(g) Cr³+(aq) + 3e¯ → Cr(s) -0.73

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Q: What is the electrical current, in Amperes, that must be applied if 6.50 g Ag is deposited at the…

Q: Physical constants that you may need: R = 8.31446 mol K C Faraday's constant F= 96485 mol e Gas…

A: Using Nernst's equation, pH of the solution can be determined from Ecell.

Q: Calculate the standard cell potential for the reaction below, and note whether the reaction is…

A: Given, Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s) Ni2+(aq)+2e−⟶Ni(s) −0.257 Mg2+(aq)+2e−⟶Mg(s) −2.372

Q: Consider the following cell notation : Cr (s) |Cr^ 3+ (aq)||H 3 AsO 4(aq) ,H^ * (aq),H 3 AsO 3(aq)…

A: In the cell notation, the left most part indicates anode. At anode oxidation takes place. After the…

Q: Calculate the cell potential for the galvanic cell in which the reaction Fe(s) + Au³*(aq) = Fe*(aq)…

Q: A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: Cu2+…

A: Step 1:

Q: Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C,…

A: The cell potential is the difference between electrode potentials of two half cells. The oxidation…

Q: 2(g) |H '(aq) || Pb<'(aq) | PBS04(s) | Pb(s) he following standard reduction potentials. q) + 2 e¯ ®…

A: Cell equation, Sn2+ + Cu2+ ---> Sn4+ + Cu Cell potential is affected by concentration. As the…

Q: Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is…

A: Given : Table of half reactions and reduction potential. To find: Complete the given table.…

Q: Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C,…

A: Use Nerenst equation and standard data E (cathode) =0.53 V E (Anode) =- 0.15 V

Q: Predict the standard potential for the galvanic cell. Refer to the table of standard reduction…

Q: (c) (c) The magnitudes but not the signs of the standard electrode potentials of two metals, X and Y…

A: Given half-cell reactions are X2+ + 2e- →X; E0 = 0.25 V Y2+ + 2e- → Y; E0 = 0.34 V Also given…

Q: Calculate the standard free-energy change for the reaction at 25 °C. Refer to the list of standard…

Q: A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2+ 2+ Sn"…

A: The reaction given is Hence in the above reaction, Sn2+ is being reduced to Sn and Ba is being…

Q: Given the reaction 4 Al(s) + 3O2(g) ===> 2Al2O3(s), write the half reactions and calculate…

A: The reaction given is, => 4 Al (s) + 3 O2 (g) --------> 2 Al2O3 (s)

Q: A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2MnO4 (aq)…

Q: Consider the battery represented by the following notation. Ni(s) | Ni*"(aq) || Ag*(aq) | Ag(s)…

Q: Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C,…

A: Answer :----- Given :---- [Cd2+ ] = 0.00100 M[Au3+ ] = 0.784 MThe overall balanced reaction are…

Q: Calculate the maximum work available (in kJ) from 52.3 g of aluminum in the following cell when the…

Q: he observed cell potential for a voltaic cell is 2.067 V when the temperature is 298 K and the…

A: Electrochemistry is the branch of chemistry which studies the change of chemical energy into…

Q: Calculate the equilibrium constant K for the following reaction at 25°C from standard electrode…

Q: Calculate the standard cell potential, Ecell, for the equation Fe(s) + F₂ (g) →→→ Fe² ¹(aq) + 2…

Q: Calculate the standard Gibbs free energy change, AGº, in kJ for the following reaction occurring in…

Q: Fe (s) + 2H+ (aq) → Fe²+ (aq) + H₂(g) Express your answer in volts to two decimal places. E cell =…

A: We have to calculate the standard cell potentials of the redox couples.

Q: A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard…

Q: Calculate the cell potential for the galvanic cell in which the reaction Fe(s) + Au³+(aq) =…

Q: Calculate the standard cell potential, E°cell, for the following reaction, and indicate if the…

A: Cu2+ (aq) + Mn2+ (aq) + 2H2O (l) ⇌Cu (s) + MnO2 (s) + 4 H+ (aq)

Q: The standard cell potential for reaction, Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) is E⁰ = 0.17V. What is…

A: The cell reaction given is, => Fe (s) + Cu2+ (aq) -------> Fe2+ (aq) + Cu (s) Given : Standard…

Q: Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C,…

Q: onsider the following balanced equation for a redox reaction in an electrochemi ell: 3 Cd (s) + 2…

Q: Calculate Ecell for the following electrochemical cell at 25 °C Pt(s) | Fe3+(aq, 0.100 M), Fe2+(aq,…

A: Answer:- This question is answered by using the simple concept of calculation of cell potential…

Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

Consider the table of the standard electrode potentials at 25 °C:
Reduction half-reaction
E° (V)
Au
13+ (aq) + 3e¯ → Au(s)
1.50
I03 (aq) + 6H*(aq) + 5e¯ → I2(aq) + 3H2O(1)
1.20
2H+(aq) + 2e → H2(g)
Cr³+ (aq) + 3e → Cr(s)
-0.73

Identify whether Au or Cr or both or neither will dissolve in 1 M HIO3 .
Both Au and Cr will dissolve.
Only Cr will dissolve.
Only Au will dissolve.
Neither Au nor Cr will dissolve.
Submit
Previous Answers
Correct
HIO3 oxidizes metals through the following reduction half-reaction:
I03 (aq) + 6H*(aq) + 5e →I2(aq) + 3H2O(1)
Because this half-reaction is above the reduction of H+ in the table, HIO3 can oxidize
some metals (for example, copper or silver) that cannot be oxidized by HCl. In general,
the metals with reduction half-reactions listed below the reduction of IO3- to I2 in the
table dissolve in HIO3 , whereas the metals listed above it do not. Therefore, Au
(which has a reduction potential of 1.50 V) will not be oxidized, but Cr (which has a
reduction potential of –0.73 V) will be oxidized by both HCl and HIO3 .
Part B
Enter a balanced redox equation for the reaction that occurs.
Express your answer as a balanced net ionic equation including phases.
ΑΣφ

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2Cr³+ (aq) + 3 Ca(s) → 2Cr(s) + 3Ca²+ (aq) Suppose the cell is prepared with 6.35 M Cr³+ in one half-cell and 0.492 M Ca²+ in the other half. Calculate the cell voltage under these conditions. Cr³+ (aq) + 3e → Cr(s) Ca²+ (aq) + 2e →→ Ca(s) There are 6 galvanic cell. Er The standard cell potential (E°) for this galvanic cell is +2.124 red = -0.744 V Ered= -2.868 V volatage is 2.15 Under these conditions, the reaction quotient is moles of electrons transferred during to operate this V V. 0.00295 In order to increase the cell voltage for this cell ✔ [ Select ] V. and the cell It is impossible to change the voltage Not enough information Q>1 Q<1 3
Identify the CORRECT cell notation for the galvanic cell (diagram below) that involves an acid-catalysed redox reaction (spectator ions are not included). V Anode Cathode Pt(s) Cu(s) +H CI Cu²+ CIO, 3 ª. Pt(s) | CIO3¯(aq), Cl¯(aq), H*(aq) || Cu(s) | Cu²+ (aq) Ob. ´Cu(s) | Cu²+ (aq) || CIO3¯(aq), CÃ¯¯(aq), H*(aq) | Pt(s) C. 2+ Pt(s) | CIO3¯(aq), CI¯(aq) || Cu(s) | Cu²+ (aq) O d. Cu(s) | Cu²+( ²+(aq) || CIO3¯¯(aq), Cl¯¯(aq) | Pt(s) e. ´Cu(s) | Cu²+ (aq) || CIO3¯(aq), CÃ¯¯(aq), H*(aq)
Calculate the cell potential of a cell operating with the following reaction at 25°C, in which [Cr2O72− ] = 0.030 M, [I − ] = 0.020, M, [Cr3+ ] = 0.30 M, and [H+ ] = 0.50 M. Cr2O72-(aq) + 6 I −(aq) + 14 H+(aq) → 2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)
Calculate the cell potential, at 298 K, for the following reaction under the conditions given.Pb(s) + Fe2+(aq, 0.0100 M) → Pb2+(aq, 0.150 M) + Fe(s) Pb2+(aq) + 2 e‒ → Pb(s) E°red = ‒0.13 VFe2+(aq) + 2 e‒ → Fe(s) E°red = ‒0.44 V Group of answer choices: ‒0.75 V ‒0.80 V ‒0.34 V ‒0.27 V ‒0.44 V ‒0.50 V
Calculate the cell potential for the galvanic cell in which the reaction Al(s) + Au³(aq) Al³(aq) + Au(s) occurs at 25 °C, given that [A1³] = 0.00140 M and [Au³+] = 0.848 M. Refer to the table of standard reduction potentials. E=
Calculate the cell potential for the galvanic cell in which the reaction Ti(s) + Cu²+ (aq) = Ti²+ (aq) + Cu(s) 2+ 2+ occurs at 25 °C, given that [Ti 1= 0.00130 M and [Cu]=0.892 M. Refer to the table of standard reduction potentials. E=
Calculate the equilibrium constant, K, for the reaction shown at 25 °C. Fe+(aq)+B(s) + 6 H, O(1) → Fe(s) + H,BO,(s) + 3 H,O*(aq) The balanced reduction half-reactions for the equation and their respective standard reduction potential values (E") are Fe3+ (aq) + 3 e- → Fe(s) H,BO, (s) + 3 H,0*(aq) + 3 e- – B(s) + 6 H, O(1) E' = -0.04 V E' = -0.8698 V K =
Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V