Question

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant...

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: Co2+(aq) + 2Cu+(aq) Co(s) + 2Cu2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values you may need in this question. Equilibrium constant: G° for this reaction would be than zero.

Homework Answers

Answer #1

Co2+(aq) + 2 Cu+ (aq) --------------> Co(s) + 2 Cu2+(aq)

Eocell = Eored - Eooxidation

          = - 0.28 - (0.153 )

          = - 0.433 V

delta Go = - n F Eocell

               = - 2 x 96485 x - 0.433

               = 83556.0 J

delta Go = - R T ln K

83556 = - 8.314 x 298 x ln K

K = 2.26 x 10^-15

equilibrium constant = 2.26 x 10^-15

G° for this reaction would be greater than zero.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1. Use standard reduction potentials to calculate the standard free energy change in kJ for the...
1. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 3I2(s) + 2Cr(s) -----> 6I-(aq) + 2Cr3+(aq) Answer: _______ kJ K for this reaction would be greater or less than one. 2. Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu2+(aq) + Sn(s) ----> 2Cu+(aq) + Sn2+(aq) Answer: ______ kJ K for this reaction would be greater or less than one.
1) The free energy change for the following reaction at 25 °C, when [Pb2+] = 1.18...
1) The free energy change for the following reaction at 25 °C, when [Pb2+] = 1.18 M and [Cd2+] = 7.90×10-3 M, is -65.9 kJ: Pb2+(1.18 M) + Cd(s)> Pb(s) + Cd2+(7.90×10-3 M) ΔG = -65.9 kJ What is the cell potential for the reaction as written under these conditions? Answer: ___V Would this reaction be spontaneous in the forward or the reverse direction? 2) Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard...
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard...
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction: Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) Fe2+(aq)+2e−→Fe(s) −0.45 Ni2+(aq)+2e−→Ni(s) −0.26
Use the table 'Standard Reduction Potentials' located in the 'Tables', to predict if a reaction will...
Use the table 'Standard Reduction Potentials' located in the 'Tables', to predict if a reaction will occur between Mn metal and Cl2(g), when the two are brought in contact via half-cells in a voltaic cell. If a reaction will occur, write a balanced net ionic equation for the reaction, assuming that the product ions are in aqueous solution. If no reaction will occur, leave all boxes blank. + +
Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) Standard Reduction (Electrode) Potentials at...
Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) F2(g) + 2 e- 2 F-(aq) 2.87 Ce4+(aq) + e- Ce3+(aq) 1.61 MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) 1.51 Cl2(g) + 2 e- 2 Cl-(aq) 1.36 Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) 1.33 O2(g) + 4 H+(aq) + 4 e- 2 H2O(l) 1.229 Br2(l) + 2 e-...
Calculate the equilibrium constant for each of the reactions at 25 ?C. Standard Electrode Potentials at...
Calculate the equilibrium constant for each of the reactions at 25 ?C. Standard Electrode Potentials at 25 ?C Reduction Half-Reaction E?(V) Pb2+(aq)+2e? ?Pb(s) -0.13 Mg2+(aq)+2e? ?Mg(s) -2.37 Br2(l)+2e? ?2Br?(aq) 1.09 Cl2(g)+2e? ?2Cl?(aq) 1.36 MnO2(s)+4H+(aq)+2e? ?Mn2+(aq)+2H2O(l) 1.21 Cu2+(aq)+2e? ?Cu(s) 0.16 Part A: Pb2+(aq)+Mg(s)?Pb(s)+Mg2+(aq) Express your answer using three significant figures. Part B: Br2(l)+2Cl?(aq)?2Br?(aq)+Cl2(g) Express your answer using two significant figures. Part C: MnO2(s)+4H+(aq)+Cu(s)?Mn2+(aq)+2H2O(l)+Cu2+(aq) Express your answer using two significant figures.
1. NH4Cl(aq)<->NH3(g) + HCl(aq) Using the standard thermodynamic data in the tables linked above, calculate the...
1. NH4Cl(aq)<->NH3(g) + HCl(aq) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. 2. HCl(g) + NH3(g)NH4Cl(s) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.
use the appropriate standard reduction potentials to determine the equlibrium constant at 200.00 K for the...
use the appropriate standard reduction potentials to determine the equlibrium constant at 200.00 K for the following reaction under acidic conditions: 4H+(aq) + MnO2(s) + 2F2+(aq) ------> Mn2+(aq) + 2Fe3+(aq) + 2H2O(l)
Part A Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Standard Reduction Potentials...
Part A Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Standard Reduction Potentials at 25 ∘C Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V Ag+(aq)+e−→Ag(s) E∘= 0.80 V
Use tabulated electrode potentials to calculate the equilibrium constant (K) at 25∘C for the following reaction....
Use tabulated electrode potentials to calculate the equilibrium constant (K) at 25∘C for the following reaction. Answer in units of x1017 Zn (s) + Ni2+ (aq) Zn2+ (aq) + Ni (s)