An 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 ina solution of 5.0 MNH3 that is also 0.00oSOM in Cu(NH3),+. what is the cell potential at 25°C?Cu?+ (aq) + 4 NH3(aq) = Cu(NH3),²+ (ag) K= 1.0 × 1013(This problem requires values in your textbook's specific appendices, which you can access throughthe OWLV2 MindTap Reader. You should not use the OWLV2 References' Tables to answer thisquestion as the values will not match.)Cell potential =VCu2*(aq) + 2 e Cu(s)0.337Cu2*(aq) + e -→ Cu*(aq)0.153

A numerical problem based on Nernst equation, which is to be accomplished.

An 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 MNH3 that is also 0.0080M in Cu(NH3), 2+ what is the cell potential at 25°C? Cu** (ag) + 4 NH3 (ag) = Cu(NH3),* (aq) K= 1.0 x 1013 (This problem requires values in your textbook's specific appendices, which you can access through the OWLV2 MindTap Reader. You should not use the OWLV2 References' Tables to answer this question as the values will not match.) Cell potential =| Cu²*(aq) + 2 e – Cu(s) 0.337 Cu2*(aq) + e– Cu*(aq) 0.153

An 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 MNH3 that is also 0.0080M in Cu(NH3), 2+ what is the cell potential at 25°C? Cu** (ag) + 4 NH3 (ag) = Cu(NH3),* (aq) K= 1.0 x 1013 (This problem requires values in your textbook's specific appendices, which you can access through the OWLV2 MindTap Reader. You should not use the OWLV2 References' Tables to answer this question as the values will not match.) Cell potential =| Cu²(aq) + 2 e – Cu(s) 0.337 Cu2(aq) + e– Cu*(aq) 0.153

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

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 120AE

See similar textbooks

Similar questions

Calculate the standard cell potential of the following cell at 25C. Sn(s)Sn2+(aq)I2(aq)I(aq)
What 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)?
At 298 K, the solubility product constant for solid Ba(IO3)2 is 1.5 109. Use the standard reduction potential of Ba2+(aq) to find the standard potential for the half-reaction Ba(IO3)2(s)+2eBa(s)+2IO3(aq)
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.
What is the standard cell potential you would obtain from a cell at 25C using an electrode in which I(aq) is in contact with I2(s) and an electrode in which a chromium strip dips into a solution of Cr3(aq)?
What is the cell potential (Ecell) of a spontaneous cell that is run at 25C and contains [Cr3+] = 0.10 M and [Ag+] = 1.0 104 M?
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.
The cell potential of the following cell at 25C is 0.480 V. ZnZn2+(1M)H+(testsolution)H2(1atm)Pt What is the pH of the test solution?
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+.