Balance the following redox equations by the ion-electron method:
1) H2O2 + Fe2+ ---> Fe3+ + H2O (in acidic solution)
CN- + MnO4- ---> CNO- + MnO2 (in basic solution)
2) Calculate the standard emf of a cell that uses the Mg/Mg2+ and Cu/Cu2+ half cell reactions at 25 deg C. What is the equation for the cell reaction that occurs under standard-state conditions?
3) If 2.50 g of CuSO4 are dissolved in 9.0 x 102 mL of 0.30 M NH3, what are the concentrations of Cu2+, Cu(NH3)2+4, and NH3 at equilibrium? (I am so lost on this one!)
4) In the complex ion [Fe(CN)6]4-, the oxidation number of Fe is_____??
an Al(s) electrode in 1.0 M Al(NO3)3 solution
a Pb(s) electrode in 1.0 M Pb(NO3)2 solution
What is the balanced overall (net) cell reaction?
a. Pb(s) + Al3+(aq) Pb2+(aq) + Al(s)
b. 3Pb(s) + 2Al3+(aq) 3Pb2+(aq) + 2Al(s)
c. 3Pb2+(aq) + 2Al(s) 3Pb(s) + 2Al3+(aq)
d. Pb2+(aq) + Al(s) Pb(s) + Al3+(aq)
6) Calculate the cell emf for the following reaction:
2Ag+(0.010 M) + H2(1 atm) 2Ag(s) + 2H+ (pH = 10.0)
7) Calculate the value of Eocell for the following reaction:
2Au(s) + 3Ca2+(aq) 2Au3+(aq) + 3Ca(s)
I don't know if seeing how I am attempting to do the problems will help in giving me some guidance, but I included the solution I reached with this problem. My question is-does a negative free energy mean I need to use a positive figure in my calculation? I am not good at algebra (you could probably tell), but I suspect this to be the case. Not mentioned in the text-I know I am supposed to know this. Your detailed explanations are a big help!
8) In the Mond process for the purification of nickel, CO is passed over the metallic nickel to give Ni(CO)4:
Ni(s) + 4CO(g)  Ni(CO)4(g)
Given that the standard free energies of formation of CO(g) and Ni(CO)4(g) are -137.3 kJ/mol and -587.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 80 Degrees C. (Assume Gfo to be independent of temperature)
Here is my attempt:
Gibbs free energy = standard free energy of formation of products - standard free energy of formation of reactants
Gibbs free energy = Gf(Ni(CO)4) - Gf(Ni(s)) -4Gf(CO)
Gibbs free energy = -587.4 - 0 + (4 x 137.3)
Gibbs free energy = -587. 4 + 549.2 = -38.2 kJ/mol
K = e (-Gibbs free energy/RT)
K = e (-38200 J/mol/(8.314 J/K.mol x 353K))
K = e (-13.016033)
= 2.2 x 10 -6 (if positive number used, would be K = e (13.016033) = 449,563.7241
9) Which is the systematic name for the compound represented below?
This solution explains a variety of problems dealing with:
1) Balancing redox reactions.
2) Calculating standard potential of a cell.
3) Calculating equilibrium concentrations.
4) Writing a net cell reaction.
5) Using free energies to calculate an equilibrium constant.