Using the standard reduction potentials listed in Appendix E in the textbook, ca

Question

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.
Part A Part complete
Fe(s)+Ni2+(aq)?Fe2+(aq)+Ni(s)
Express your answer using two significant figures.
K =   
2.6×10^5
Correct Answer

Part B
Co(s)+2H+(aq)?Co2+(aq)+H2(g)
Express your answer using two significant figures.

K =   

Part C
10Br?(aq)+2MnO?4(aq)+16H+(aq)?2Mn2+(aq)+8H2O(l)+5Br2(l)
Express your answer using two significant figures.

K =

Explanation / Answer

ANSWER:

(a) Fe(s)+Ni2+(aq)?Fe2+(aq)+Ni(s)

Eo = EoNi2+/Ni - EoFe2+/Fe ; Ni2+ undergoes reduction and Fe undergoes oxidation.

(-0.28V) - (-0.44V) = 0.18V

we know

Eo = 0.0591 /n logK ' n= number of electrons involved in balanced reaction = 2

logK = n  Eo / 0.0591 = 0.168 X 2 / 0.0591 = 5.68

K = antilog(5.68) = 4.78 X 105

(NOTE: You can insert the EoNi2+/Ni and EoFe2+/Fe values given appendex E to get exactly 2.6 X 105)

(b)

Co(s)+2H+(aq)?Co2+(aq)+H2(g)

Eo = EoH2+/H2 - EoCo2+/Co = (0.00V) - (-0.277V) = 0.277V

K = antilog(0.277 X 2) / 0.0591 = 4.78 X 107

(c)

10Br?(aq)+2MnO4?(aq)+16H+(aq)?2Mn2+(aq)+8H2O(l)+5Br2(l)

Eo = - EoMn2+/MnO4? - EoBr/Br- = 1.5119 - 1.065 = 0.4469V

K = antilog(5 X 0.4469 / 0.0591) = 6.30 X 1037

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