Part A
Calculate the equilibrium constant at 25 ∘C for the
reaction
Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s),
if
Ni2+(aq)+2e−→Ni(s), E∘
= -0.26 V,
Al+(aq)+e−→Al(s), E∘ =
0.80 V
Express your answer using one significant figure.
Part B
Calculate the equilibrium constant at 25 ∘C for the
reaction
Hg2+2(aq)→Hg(l)+Hg2+(aq)
See Appendix D for standard reduction potentials.
Express your answer using one significant figure.
(a) Calculate the equilibrium constant for the following
reaction at 25°C: Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq) The standard
cell potential of the corresponding voltaic cell is 0.030 V. (b)
When equal volumes of 1.3 M solutions of Ag+ and Fe2+ are mixed,
what is the equilibrium concentration of Fe2+? Ag+(aq) + Fe2+(aq)
Ag(s) + Fe3+(aq)
Calculate the equilibrium constant for each of the reactions at
25 ∘C .
A.)
2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq)
B.)
O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)
C.)
Br2(l)+2I−(aq)→2Br−(aq)+I2(s)
4. Calculate the equilibrium constant at standard temperature,
25oC, for the redox reaction between Pb2+(aq) and Zn(s). HINT:
Write a balanced chemical reaction first. a) 1.17x1030 b) 1.92x1021
c) 4.38x1010 d) 1.08x1015 e) 1.75x109
What is the value of the equilibrium constant for the reaction
of Na(s)with Pb2+ (aq)at 25° C? Eº cell = 2.58 V 2Na(s) + Pb2+(aq)
↔ Pb(s) + 2Na+(aq) 3.0 x 1043 6.0 x 10200 6.0 x 10 43 1.8 x
1087
Under standard conditions, Consider the following standard
reduction potentials,
Ni2+(aq) + 2 e- → Ni(s) E° = -0.26 V I2(s) + 2 e- → 2 I-(aq) E°
= +0.54 V Under standard conditions,
a)Ni2+(aq) is a stronger oxidizing agent than I2(s) and I-(aq)
is a stronger reducing agent than Ni(s).
b)I2(s) is a stronger oxidizing agent than Ni2+(aq) and Ni(s) is
a stronger reducing agent than I-(aq).
c)I-(aq) is a stronger oxidizing agent than Ni(s) and I2(s) is a
stronger...
Calculate ΔG° for the following reaction as written
Ni2+ (aq) + Sn (s) → Ni (s) + Sn2+
(aq)
given the following reduction half-reactions and standard
reduction potentials.
Sn2+ (aq) + 2 e- → Sn (s)
E° = -0.15V
Ni2+ (aq) + 2 e- → Ni (s)
E° = -0.25V