From the observations listed, estimate the value of
a. The metal M reacts with
b. The metal M reacts with HCI(aq), producing
(a)
Interpretation:
The value of E°cellshould be estimated for the half reaction of
Concept introduction:
The solid metal should have higher reactivity than the aqueous metal in order to observe the displacement.
In order to react with an acid, the acid anion should have higher reduction potential while oxidizing the metal to metal cation in the acidic medium.
Answer to Problem 1E
Explanation of Solution
If M displaces the Ag+(aq) the
Therefore, lower limit of the estimation for the
(b)
Interpretation:
The value of
Concept introduction:
The solid metal should have higher reactivity than the aqueous metal in order to observe the displacement.
In order to react with an acid, the acid anion should have higher reduction potential while oxidizing the metal to metal cation in the acidic medium.
Answer to Problem 1E
Explanation of Solution
If displaces neither Zn+(aq) nor Fe2+(aq), the
Therefore, lower limit of the estimation for the
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Chapter 19 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
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- You have 1.0 M solutions of Al(NO3)3 and AgNO3 along with Al and Ag electrodes to construct a voltaic cell. The salt bridge contains a saturated solution of KCl. Complete the picture associated with this problem by a writing the symbols of the elements and ions in the appropriate areas (both solutions and electrodes). b identifying the anode and cathode. c indicating the direction of electron flow through the external circuit. d indicating the cell potential (assume standard conditions, with no current flowing). e writing the appropriate half-reaction under each of the containers. f indicating the direction of ion flow in the salt bridge. g identifying the species undergoing oxidation and reduction. h writing the balanced overall reaction for the cell.arrow_forwardFrom the information provided, use cell notation to describe the following systems: (a) In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-Cell, Cu metal goes into a.Cu(NO3)2 solution with all solute concentrations 1 M. (b) The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution. (c) One half-cell consists of a silver electrode in a 1 M AgNO3 solution, and in the other half-cell, a copper Electrode in 1 M Cu(NO3)2 is oxidized.arrow_forwardGiven this reaction, its standard potential, and the standard half-cell potential of 0.34 V for the Cu2+ |Cu half-cell, calculate E° for the Fe(s)|Fe2+(aq) half-cell.arrow_forward
- It took 150. s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of 1+.arrow_forwardCalculate the standard cell potential of the cell corresponding to the oxidation of oxalic acid, H2C2O4, by permanganate ion. MnO4. 5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l) See Appendix C for free energies of formation: Gf for H2C2O4(aq) is 698 kJ.arrow_forward
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