Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions:
(a) Cl2 (g) + 2I- (aq)→ 2Cl- (aq)+ I2(s)
(b) Ni(s) + 2 Ce4+(aq) →Ni2+(aq) + 2 Ce3+(aq)
(c) Fe(s) + 2 Fe3+(aq)→3 Fe2+(aq)
(d) 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)→ 2NO(q) + 4H2O(l) + 3 Cu2+(aq)
One Class Solution:-
a)CI2(g) +2e-→2Cl-(aq) Eo = 1.359 V
2I- (aq)→+ I2(s)+ 2e- Eo = 0.536 V, E0= 1.359-0.536=0.823V
b) Ni(s) →Ni2+(aq) + 2e- anode Eo = –0.28V
2[Ce4+(aq) +e- →2 Ce3+(aq)] cathode Eo =1.61 V
Eo = 1.61-(-0.28)=1.89 V
c) Fe(s) + 2 Fe3+(aq)→3 Fe2+(aq) Eo =?
Fe(s) → Fe2+(aq) +2e- Eo = -0.440V
Fe3+(aq) +e- → Fe2+(aq) Eo = 0.771V
E=0.771-(-0.440)= 1.211 V
(d) 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)→ 2NO(q) + 4H2O(l) + 3 Cu2+(aq)
2 NO3-(aq) + 8 H+(aq) + 3e-→ 2NO(q) + 4H2O(l) Eo = 0.96V
Cu(s)→ Cu2+(aq) +2e- Eo = V+0.377
Eo =0.96-0.377= 0.583V
Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions:
(a) Cl2 (g) + 2I- (aq)→ 2Cl- (aq)+ I2(s)
(b) Ni(s) + 2 Ce4+(aq) →Ni2+(aq) + 2 Ce3+(aq)
(c) Fe(s) + 2 Fe3+(aq)→3 Fe2+(aq)
(d) 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)→ 2NO(q) + 4H2O(l) + 3 Cu2+(aq)
One Class Solution:-
a)CI2(g) +2e-→2Cl-(aq) Eo = 1.359 V
2I- (aq)→+ I2(s)+ 2e- Eo = 0.536 V, E0= 1.359-0.536=0.823V
b) Ni(s) →Ni2+(aq) + 2e- anode Eo = –0.28V
2[Ce4+(aq) +e- →2 Ce3+(aq)] cathode Eo =1.61 V
Eo = 1.61-(-0.28)=1.89 V
c) Fe(s) + 2 Fe3+(aq)→3 Fe2+(aq) Eo =?
Fe(s) → Fe2+(aq) +2e- Eo = -0.440V
Fe3+(aq) +e- → Fe2+(aq) Eo = 0.771V
E=0.771-(-0.440)= 1.211 V
(d) 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s)→ 2NO(q) + 4H2O(l) + 3 Cu2+(aq)
2 NO3-(aq) + 8 H+(aq) + 3e-→ 2NO(q) + 4H2O(l) Eo = 0.96V
Cu(s)→ Cu2+(aq) +2e- Eo = V+0.377
Eo =0.96-0.377= 0.583V
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