Concept explainers
Interpretation:
Time needed to cover the object with gold to a thickness of
Concept Introduction:
The Faraday’s first law of
Here,
The symbol
The symbol
The symbol
In the simple case of constant current electrolysis,
Here,
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Chemistry: Principles and Practice
- An electrolysis experiment is performed to determine the value of the Faraday constant (number of coulombs per mole of electrons). In this experiment, 28.8 g of gold is plated out from a AuCN solution by running an electrolytic cell for two hours with a current of 2.00 A. What is the experimental value obtained for the Faraday Constant?arrow_forwardA voltaic cell is constructed in which one half-cell consists of a silver wire in an aqueous solution of AgNO3.The other half cell consists of an inert platinum wire in an aqueous solution containing Fe2+(aq) and Fe3+(aq). (a) Calculate the cell potential, assuming standard conditions. (b) Write the net ionic equation for the reaction occurring in the cell. (c) Which electrode is the anode and which is the cathode? (d) If [Ag+] is 0.10 M, and [Fe2+] and [Fe3+] are both 1.0 M, what is the cell potential? Is the net cell reaction still that used in part (a)? If not, what is the net reaction under the new conditions?arrow_forwardYou 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_forward
- Calculate 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_forwardCalculate the cell potential of a cell operating with the following reaction at 25C, in which [MnO4] = 0.010 M, [Br] = 0.010 M. [Mn2] = 0.15 M, and [H] = 1.0 M. 2MNO4(aq)+10Br(aq)+16H+(aq)2MN2(aq)+5Br2(l)+8H2O(l)arrow_forwardThe mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).arrow_forward
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