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

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) Ered = -0.744 V Ca2+ (aq) + 2e → Ca(s) Ered = -2.868 V The standard cell potential (E°) for this galvanic cell is [Select] V. There are [ Select] + moles of electrons transferred during to operate this galvanic cell. Under these conditions, the reaction quotient is [Select] and the cell volatage is [Select] V. In order to increase the cell voltage for this cell, [Select ]
The reaction 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) was used in a galvanic cell. E°=0.46 V for this reaction. What is the cell potential, E, at 298K if the concentrations are [Cu2+] = 1.0M and [Ag+] = 0.10M?
Using the following standard reduction potentials E° = -2.92 V Cs*(aq) + e –→ Cs(s) Cr2*(aq) + 2 e » Cr(s) E° = -0.91 V calculate the standard cell potential and write the cell reaction for a voltaic cell. E°cell = -2.01 V and Cr2+(aq) + 2 Cs(s) → 2 Cs*(aq) + Cr(s) E°c cell = +2.01 V and Cr2*(aq) + 2 Cs(s) → 2 Cs*(ag) + Cr(s) None of these E°cell = +4.91 V and %3D Cr(s) + 2 Cs*(aq) →2 Cs(s) + Cr2+(aq) E°cell = -4.91 V and Cr(s) + 2 Cs*(aq) → 2 Cs(s) + Cr2+(aq)
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=
Consider the following redox reaction. Al (s) + Cr3+ (aq) → Al3+ (aq) + Cr (s) Calculate the standard cell potential (E°cell) for the reaction (V).
Using the Table of Standard Electrode Potentials (Appendix L, page 1251), calculate the E°cell for each of the following reactions and predict whether each redox reaction is spontaneous or non-spontaneous. 2. 3+, (a) Au(s) + NO; (aq) + 4H*(aq) → Au³*(aq) + NO(g) + 2H2O(); (b) Сu's) + 2Fe?" (ag) —> Сu?"(aq) + Fe(s) ; (c) 2Fe" (ag) + 21 (aд) —> I2(ад) + 2Fe?"(aq); 2+ (d) Zn(OH)2(s) + 4NH3(aq) → Zn(NH3)4*(aq) + 20H (aq);
For a galvanic cell based on the following half-reactions: Fe3+(aq) + e- → Fe2+(aq)Ag+(aq) + e- → Ag(s)
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
For the cell shown, the measured cell potential, Ecell, is −0.3627 V at 25 °C. Pt(s) | H2(g,0.735 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, ?o, are 2H+(aq)+2e−⟶H2(g) ?o=0.00 V Cd2+(aq)+2e−⟶Cd(s) ?o=−0.403 V Calculate the H+ concentration in M.