If the standard reference half-cell used construct a table of electrode potentials was Zn²+ (aq) + 2e - Zn(s), the standard electrode potential for Fe²+ (aq) + 2e - Fe(s) would beSelect one:O +0.31VO -0.45VO +0.45VO -0.31VIf the standard reference half-cell used construct a table - electrode potentials was Zn²+ (aq) + 2e Zn(s), the cell potential for a Fe(s) | Fe²(aq) || Cu²(aq) | Cu(s) cell would beSelect one:O-0.79VO-1.55VO +1.55VO +0.79VThe cost of producing a Cu(s) | Cu2+ (aq) Au³+ (aq) | Au(s) cell can be reduced without affecting the cell voltage. Simply replace thewith a less expensive

Zn2+(aq) + 2e- → Zn(s) Eo=-0.76 VFe2+(aq) + 2e- → Fe(s) Eo=-0.45 VCu2+(aq) +…

If the standard reference half-cell used to construct a table of electrode potentials was Zn²+ (aq) + 2e - Zn(s), the standard electrode potential for Fe2(aq) + 2e - Fe(s) would be Select one: O +0.31V O-0.45V O +0.45V O-0.31V If the standard reference half-cell used to construct a table of electrode potentials was Zn² (aq) + 2e- Zn(s), the cell potential for a Fe(s) | Fe²(aq) || Cu²(aq) | Cu(s) cell would be Select one: O-0.79V O-1.55V O +1.55V O +0.79V

If the standard reference half-cell used to construct a table of electrode potentials was Zn²+ (aq) + 2e - Zn(s), the standard electrode potential for Fe2(aq) + 2e - Fe(s) would be Select one: O +0.31V O-0.45V O +0.45V O-0.31V If the standard reference half-cell used to construct a table of electrode potentials was Zn² (aq) + 2e- Zn(s), the cell potential for a Fe(s) | Fe²(aq) || Cu²(aq) | Cu(s) cell would be Select one: O-0.79V O-1.55V O +1.55V O +0.79V

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 consists of a Ni/Ni²+ half-cell and a Co/Co²+ half-cell with initial concentrations: [Ni²+] = 0.80M and [Co²+] = 0.20 M at 298.15 K. half-reaction Eº (V) Ni²+ (aq) +2e→Ni(s) -0.25 Co²+ (aq) +2e→Co(s) -0.28 Write the balanced overall reaction that powers the cell. Calculate Ecell for the redox reaction. (round to 2 decimal places) What is [Ni²+] when Ecell when reaches 0.03V? (round to 2 sig figs) V What happens at equilibrium? What is Ecell at equilibrium? [Ni²+] at equilibrium: [Co2+] at equilibrium: (round to 2 decimal places) (round to 2 sig figs) (round to 2 sig figs) V M M M
Given the following electrochemical cell, calculate the potential for the cell in which the concentration of Ag+ is 0.0765 M, the pH of the H* cell is 2.700, and the pressure for H₂ is held constant at 1 atm. The temperature is held constant at 55°C. Ag Agt H+ + Ag+le ←H₂(g) Standard reduction potentials: 2H+ + 2e H₂ Fº = 0 V → Ag E°=0.800 V
When [Hg2+] = 1.16 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 2.583 V. What is the Al3+ concentration in this cell? 3H92*(aq) + 2Al(s) – 3Hg(8) + 2AI3*(aq) [A|3*] = mol/L 6.724x10^-4 Show Approach Hide Tutor Steps Resubmit TUTOR STEP The first step is to separate the overall reaction into oxidation and reduction half-reactions. These are shown here. Hg2*(aq) + 2e" – Hg(e) reduction Al(s) → A3*(aq) + 3e oxidation Next (1 of 5)
Cu = reduction potential of +0.52. Mg is in pic. Thanks
A galvanic (voltaic) cell consists of an inert platinum electrode in a solution containing 1.0 M1.0 M tin(IV) ion and 1.0 M1.0 M tin(II) ion, and another inert platinum electrode in a solution containing 1.0 M1.0 M cerium(IV) ion and 1.0 M1.0 M cerium(III) ion, connected by a salt bridge. Calculate the standard potential for this cell at 25°C25°C. Standard reduction potentials can be found in the standard reduction potentials table.
Use an inactive electrode when needed. Consider the following cell reactions and its standard cell potential: 3 Hg22+(aq) + 2 Cr3+(aq) ⟶⟶2 Cr(s) + 6 Hg2+(aq) E°cell = -1.66 V a) Write the cell notation for the voltaic cell. b) What are the stoichiometric number of moles of electrons delivered from the one electrode compartment to the other? c) Draw a diagram of the voltaic cell. Label the anode and cathode compartments and the sign of each.Label what ions, gases, and solids are in each compartment.Label the electrode in each compartment. In other words, what is conducting electrons in and out of each compartment.Label the direction of electron flow.Label the salt bridge and what ionic compound is being used in the salt bridge. Label the direction of the ion flow for the cation and anion of the salt bridge.
What is the standard cell potential for a galvanic cell made up of Ni and Zn?
When [Pb2+] = 1.49 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 0.749 V. What is the Zn2+ concentration in this cell? Pb2+(aq) + Zn(s) → Pb(s) + Zn²+(aq) [Zn2+] = mol/L
A galvanic cell consists of a Ni/Ni²+ half-cell and a Co/Co²+ half-cell with initial concentrations: [Ni²+] = 0.80M and [Co²+] = 0.20 M at 298.15 K. half-reaction Eº (V) Ni²+ (aq) +2e→Ni(s) -0.25 Co²+ (aq) +2e→Co(s) -0.28
What is the cell potential (EO) for a galvanic cell formed from the following two half-reactions? PbO2(s) + 3 H*(a) + HSO4-(aq) + 2 e-+ PBSO4(s) + 2 E° = 1.690 V %3D H20(1) PbSO4(s) + H*(aq) + 2 e- → Pb(s) + HSO4-(aq) E° = -0.356 V (A) +1.334 V (B) -2.046 V c) +2.046 V D) +2.758 V
For the galvanic cell: Zn(s) + 2Tl+(aq) 2Tl(s) + Zn2+(aq) Eocell = 0.42 V The standard reduction potential for the half cell Zn2+(aq) + 2e– Zn(s) is Eored = –0.76 V What is the standard reduction potential for Tl+(aq) + e– Tl(s) Eored = ? V a. +0.17 V b. –0.17 V c. –0.34 V d. +0.34 V
A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode composed of copper in a 1.0 M copper(I) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C. Refer to the list of standard reduction potentials. E cell || = V