(a)
Interpretation:
The equilibrium constant of given cell should be calculated and concentration of
Concept introduction:
Nernst equation:
The relationship between standard cell potential and cell potential at non-standard conditions and the reaction quotient are given by Nernst equation it is,
Where,
(b)
Interpretation:
The equilibrium constant of given cell should be calculated and concentration of
Concept introduction:
Nernst equation:
The relationship between standard cell potential and cell potential at non-standard conditions and the reaction quotient are given by Nernst equation it is,
Where,
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General Chemistry - Standalone book (MindTap Course List)
- Another type of battery is the alkaline zinc-mercury cell, in which the cell reaction is Zn(s) + HgO(s) Hg() + ZnO(s) E = + 1.35 V (a) What is the standard free energy change for this reaction? (b) The standard free energy change in a voltaic cell is the maximum electrical energy that the cell can produce. If the reaction in a zinc-mercury cell consumes 1.00 g mercury oxide, what is the standard free energy change? (c) For how many hours could a mercury cell produce a 10-mA current if the limiting reactant is 3.50 g mercury oxide?arrow_forwardWhat is the standard cell potential you would obtain from a cell at 25C using an electrode in which Hg22+(aq) is in contact with mercury metal and an electrode in which an aluminum strip dips into a solution of Al3+(aq)?arrow_forwardFour voltaic cells are set up. In each, one half-cell contains a standard hydrogen electrode. The second half-cell is one of the following: (i) Cr3+(aq, 1.0 M)|Cr(s) (ii) Fea+(aq, 1.0M)|Fe(s) (iii) Cu2+(aq, 1.0M)|Cu(s) (iv) Mg2+(aq, 1.0M)|Mg(s) (a) In which of the voltaic cells does the hydrogen electrode serve as the cathode? (b) Which voltaic cell produces the highest potential? Which produces the lowest potential?arrow_forward
- At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction PbC2O4(s)+2ePb(s)+C2O42(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/PbC2O4 electrode in a 0.025 M solution of Na2C2O4.arrow_forwardAt 298 K, the solubility product constant for Pb(IO3)2 is 2.6 1013, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction Pb(IO3)2(s)+2ePb(s)+2IO3(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions, and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/Pb(IO3)2 electrode in a 3.5 103 M solution of NaIO3.arrow_forwardAn electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag+] = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 M NH3 that is also 0.010 M in Cu(NH3)42+, what is the cell potential at 25C? Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)K=1.01013arrow_forward
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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_forwardIt 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_forward
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