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Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K:

  1. Fe(s) + Ni2+(aq)→ Fe2+(aq) + Ni(s)

  2. Co(s) + 2 H+(aq)→Co2+(aq) + H2(g)

  3. 10 Br-(aq) + 2 MnO4-(aq) + 16 H+(aq)→ 2 Mn2+(aq) + 8 H2O(aq) + 5 Br2(l)

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Reid Wolff
Reid WolffLv2
24 May 2020

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Related questions

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 the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K:

(a) Cu(s) + 2 Ag+(aq)→Cu2+(aq) + 2 Ag(s)

(b) 3 Ce4+(aq) + Bi(s) + H2O(l)→3Ce3+(aq) +BiO+(aq) + 2H+(aq)

(c) N2H5+(aq) + 4 Fe(CN)63-(aq)→N2(g) +5H+(aq) + 4Fe(CN)64-(aq)


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