The standard emf

1. Calculate the standard emf for each of the following reactions:

a)H2(g) + F2(g)—> 2H+(aq) + 2F-(aq)
b)Cu(s) + Ba 2+(aq)—> Cu 2+(aq) + Ba(s)
c)3Fe 2+(aq)—> Fe(s) + 2Fe 3+(aq)
d)Hg2 2+(aq) + 2Cu+(aq)—-> 2Hg(l) + 2Cu 2+(aq)

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2. The standard reduction potential for the reduction of RuO4-(aq) to RuO4 2-(aq) is +0.59 V. Which of the following substances can oxidize RuO4 2-(aq) to RuO4-(aq) under standard conditions: Cr2O7 2-(aq), ClO-(aq), Pb 2+(aq), I2(s), Ni 2+(aq)?

3. Using standard reduction potentials, calculate the equilibrium constant for each of the following reactions at 298 K:
a) 2 VO2+(aq) + 4 H+(aq) + 2 Ag(s)—–> 2 VO 2+(aq) + 2 H2O(l), 2 Ag+(aq)
b) 3 Ce 4+(aq) + Bi(s) + H2O(l) —–> 3 Ce 3+(aq) + BiO+(aq) + 2 H+(aq)
c) N2H5+(aq) + 4 Fe(CN)6 3-(aq)—–> N2(g) + 5 H+(aq) + 4 Fe(CN)6 4-(aq)

4. A voltaic cell is constructed that is based on the following reaction:
Sn 2+(aq) + Pb(s)—-> Sn(s) + Pb 2+(aq)
a) If the concentration of Sn 2+ in the cathode compartment is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb 2+ in the anode compartment?

b) If the anode compartment contains [SO4 2-]= 1.00 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?

5. a) Draw two linkage isomers of [Co(NH3)5 SCN]2+
b) Draw the two geometric isomers of [Co(NH3)3 Cl3]2+
c) Two compounds with the formula Co(NH3)5ClBr can be prepared. Use structural formulas to show how they differ. What kind of isomerism does this illustrate?

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1. Calculate the standard emf for each of the following reactions:
a)H2(g) + F2(g)—> 2H+(aq) + 2F-(aq)
H2(g) 2H+(aq) + 2e- E°red = 0.00 V
F2(g) + 2e- 2F -(aq) E°red = 2.87 V
_______________________________________________
H2(g) + F2(g) 2H+(aq) + 2F -(aq) E° = 2.87 V
E° = E°red(reduction) – E°red(oxidation) = 2.87 V – 0.00 V = 2.87 V

b)Cu(s) + Ba 2+(aq)—> Cu 2+(aq) + Ba(s)
Cu(s) Cu2+(aq) + 2e- E°red = 0.337 V
Ba2+(aq) + 2e- Ba(s) E°red = -2.90 V
___________________________________________________
Cu(s) + Ba2+(aq) Cu2+(aq) + Ba(s) E° = -3.24 V
E° = E°red(reduction) – E°red(oxidation) = -2.90 V – 0.337 V = -3.24 V

c)3Fe 2+(aq)—> Fe(s) + 2Fe 3+(aq)
Fe2+(aq) + 2e- Fe(s) E°red = -0.440 V
2Fe2+(aq) 2Fe3+(aq) + 2e E°red = 0.771 V
___________________________________________________
3Fe2+(aq) Fe(s) + 2Fe3+(aq) E° = -1.211 V
E° = E°red(reduction) – E°red(oxidation) = -0.440 V – 0.771 V = -1.211 V

d)Hg2 2+(aq) + 2Cu+(aq)—-> 2Hg(l) + 2Cu 2+(aq)
Hg22+(aq) + 2e- 2Hg(l) E°red = 0.789 V
2Cu+(aq) 2Cu2+(aq) + 2e E°red = 0.153 …

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