Use the tabulated electrode potentials to calculate K for the oxidation of iron

Question

Use the tabulated electrode potentials to calculate K for the oxidation of iron by H+:

2Fe(s)+6H+(aq)2Fe3+(aq)+3H2(g)

Express your answer using two significant figures.

Standard reduction half-cell potentials at 25C

Half-reaction E (V) Half-reaction E (V) Au3+(aq)+3eAu(s) 1.50 Fe2+(aq)+2eFe(s) 0.45 Ag+(aq)+eAg(s) 0.80 Cr3+(aq)+eCr2+(aq) 0.50 Fe3+(aq)+3eFe2+(aq) 0.77 Cr3+(aq)+3eCr(s) 0.73 Cu+(aq)+eCu(s) 0.52 Zn2+(aq)+2eZn(s) 0.76 Cu2+(aq)+2eCu(s) 0.34 Mn2+(aq)+2eMn(s) 1.18 2H+(aq)+2eH2(g) 0.00 Al3+(aq)+3eAl(s) 1.66 Fe3+(aq)+3eFe(s) 0.036 Mg2+(aq)+2eMg(s) 2.37 Pb2+(aq)+2ePb(s) 0.13 Na+(aq)+eNa(s) 2.71 Sn2+(aq)+2eSn(s) 0.14 Ca2+(aq)+2eCa(s) 2.76 Ni2+(aq)+2eNi(s) 0.23 Ba2+(aq)+2eBa(s) 2.90 Co2+(aq)+2eCo(s) 0.28 K+(aq)+eK(s) 2.92 Cd2+(aq)+2eCd(s) 0.40 Li+(aq)+eLi(s) 3.04

Explanation / Answer

reduction half reaction =>

H2(g)   --------> H2+ + 2 e-   E= 0

Oxidation half reaction =>

Fe(s) ------>   Fe3+ + 3 e-     E= -0.036 V

E0cell = Ecathode - Eanode = 0 - (-0.036)= 0.036 V

G0 =- n F E0 = - 6 *96500 * 0.036 = -20844 J

G0 = - 2.303 * R *T log K

-20844 J = -2.303 * 8.314 * 298 log K

log K = 3.6530

K= 4497.79

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