CHEMWORK Porous disk Cu Cu Cu* Cu* The concentration cell shown above employs the reaction Cu2+ + 2 e- Cu E° = 0.34 V Both compartments of this cell have copper metal electrodes. The compartment on the left contains 6.0x10-3M Cu2+ and the compartment on the right contains 7.2×102M Cu2+. The expected potential for this cell at 25°C is V

CHEMWORK Porous disk Cu Cu Cu* Cu* The concentration cell shown above employs the reaction Cu2+ + 2 e- Cu E° = 0.34 V Both compartments of this cell have copper metal electrodes. The compartment on the left contains 6.0x10-3M Cu2+ and the compartment on the right contains 7.2×102M Cu2+. The expected potential for this cell at 25°C is V

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

Consider the following half-cell reactions and associated standard half-cell potentials Ag+ + e- ? Ag Cu+2 + 2e- ? Cu Ni+2 + 2e- ? Ni Cr+3 + 3e- ? Cr Write the overall reaction using the list above that leads to the maximum positive standard potential. What is the maximum standard potential?
The Nernst Equation A nonstandard cell or half-cell potential can be calculated using the Nernst equation: RT E = E- InQ nF where E = potential under nonstandard conditions E = standard potential R = ideal gas constant 8.314 J/K mol %3D %3D T = kelvin temperature n = number of moles of electrons for the reaction as written %3D F = charge carried by 1 mol of electrons = 96, 485 C/mol = 96, 485 J/V · mol %3D %3D = reaction quotient It is customary to use the equation in a form where numberical values are substituted for R, T, and F at a temperature of 25°C. RT (8.314 J/K - mol)(298K) 96, 485 J/V mol = 0.02568 V %3D F and the Nernst equation with the potentials in volts is: E = E° – 0.02568 In Q - Sometimes it is more convenient to use base-10 logarithms, and the substitution of 2.3026 log for In is made. (2.3026 x 0.02568 = 0.0591) E = E° – 0.0591 logQ A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm. What is…
2. A galvanic cell consists of a Mn electrode in a 1.0 M Mn(NO3)2 solution and an Al electrode in a 1.0 M AI(NO3)3 Solution. Calculate Eºcell for this cell at 25°C Mn²+ + 2e → Mn Eº = -1.05 V Al³+ + 3e → Al Eº = -1.67 V
g and lea X rn/takeAssignment/takeCovalentActivity.do?locator%3Dassignment-take [Review Topics] [References] Use the References to access important values if needed for this question. What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the M and the Mn²+ concentration is 1.05 M ? 4 Pb2+ concentration is 9.1l × 10 Pb²+ (aq) + Mn(s) → Pb(s)+ Mn²+ (aq) Ecell V for the concentrations given. The cell reaction as written above is
28). What statement about a galvanic cell constructed with Cu*/Cu and Fe2+/Fe is correct? Standard Reduction Potentials E°red / V Cu* + e→ Cu +0.52 Fe2+ + 2e→ Fe -0.44 a. Fe(s) will reduce Cu*(aq). b. Fe2*(aq) is reduced at the cathode. c. The standard cell potential (E°cell) is +1.40 V. d. Fe(s) is a stronger oxidizing agent than Cu(s).
Calculate the value of the potential for the following electrochemical cell under the conditions indicated at 25ºC.Cr |Cr+3 (1.75M)∥Co+2 (0.050M)|Co
7. Calculate the Ecell value at 298 K for the cell based on the following reaction when [Fe3] = [Cu"] = 1.50 x 103 M and [Fe²*] = [Cu²*] = 2.50 x 104 M. Use the standard reduction potentials provided below %3D Fe"(aq) + Cu (aq) → Fe**(aq) + Cu*(aq)
In this aluminum concentration cell, the reaction would proceed spontaneously (from the left to the right compartment or right to left compartment)
Salt bridge U A concentration cell similar to the one shown is composed of two Cu electrodes and solutions of different Cu²+ concentrations. The left compartment contains 0.272 M Cu²+, and the right compartment contains 1.25 M Cu²+ . Calculate the cell potential for this reaction at 298 K. volts In this copper concentration cell, the compartment on the left is the + and the compartment on the right is the +
The Nernst Equation A nonstandard cell or half-cell potential can be calculated using the Nernst equation: RT InQ nF E = E° - where E = potential under nonstandard conditions %3D E = standard potential R = ideal gas constant = 8.314 J/K mol %3D T = kelvin temperature n = number of moles of electrons for the reaction as written F = charge carried by 1 mol of electrons = %3D 96, 485 C/mol = 96, 485 J/V mol %3D Q = reaction quotient %3D It is customary to use the equation in a form where numberical values are substituted for R, T, and F at a temperature of 25°C. (8.314 J/K mol)(298K) 96, 485 J/V mol RT = 0.02568 V %3D and the Nernst equation with the potentials in volts is: 0.02568 In Q E = E - n Sometimes it is more convenient to use base-10 logarithms, and the substitution of 2.3026 log for In is made. (2.3026 x 0.02568 = 0.0591) %3D 0.0591 logQ E = E° – A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm. Use…
Question 3 D Voltage Source Zn 로 salt bridge ZnSO4 Question 4 If the cell pictured above is driven as an electrolytic cell: The electrons would flow (Select ] The following statement is true: [Select] Cu The standard (assuming 1 M concentrations) cell potential is [Select ] If the salt bridge contains KCI, then [Select] CuSO4 In order to decrease (make it a smaller number) the potential necessary for electrolysis, [Select] If the current is set to a constant of 3.82 A for 5.00 min, then (Select] of metal would deposit on the cathode.
Q4 A: What is the E for the Fe3+/Fe2+ half-cell if [Fe3+] = 0.001 M and [Fe2+] is 0.1 M at 25 °C? if you know Fe reduction potential = + 0.771 V *