Use data from Table 20.1 to find the standard reduction potential for (3) Fe3+(a

Question

Use data from Table 20.1 to find the standard reduction potential for (3) Fe3+(aq) + 3e Fe(s) E3°= ?

Potentials at 25 C Reduction Half-Reactian Acidic solution F2(g) 2e 2 F (aq +2.866 oa(g) 2 H (aq) 2 e O2 (g) H2O(l) 2.075 S203 (aq) 2 +2.01 H202(aq) 2 H (aq) 2e 2H20 +1,763 Mn2 (aq) 4H20 (1) Mnol (aq) 8 H (aq 5e +1.51 Pbo.(s) 4H (aq) 2 e +1.455 2e +1.358 Cryo? (aq) 14H (aq) 6 e 2Crs (aa) 7H200) +1.33 Mno2(s) 4H (aq) 2e aq) 2H20 (1) +1.23 o2(g) +4H (aq) 4e 2H20 +1.229 210, (aq) 12 HI (aq) 10e +1.20 2e 2 Br (aq) 1.065 Noa (aq) 4H (aq) 3e NO (g) 2H20() +0.956 Ag (s) +0.800 8 (aq) e Feu (aq) e +0,771 2 H (aq) 2e H202(aq) I2(s) 2e +0,538 Cu Haq) 2 Cu (s) +0,340 SO, (aq) 4 H (aq) 2, e +0.17 aq) 2 e +0.154 s(s) 2H (aq) 2 e H2S(8) +0.14 2 H (aq) 2 e Hz(g) Pb2 (aq) 2 Pb(s) -0.125 Sn(s 2 0.137 Fe" (aq) 2 e Fe(s) aq) 2 e Zn(s) AP (aq) 3 e Al(s) Mg2 (aq) 2 e Mg(s) 2.356 Na (aq) e Nats) -2.713 Caz (aq) 2 e Ca(s -2.84 K (aq) K(s) 2.924 Li (aq) e Li(s 3.0N0 Basic solution o30g) H20 (l) 2e 020g) 20H (aq) +1.246 OCI (aq) H20 2e Cl (aq) 2 OH (aq +0.890 o20g) 2H20 (I) 4 4 OH (aq) +0.101 2H20 (l) 2e H2(g) 20H (aq

Explanation / Answer

from the data table we have:

Fe3+ (aq) + e- —> Fe2+ (aq) … eqn 1

Eo = +0.771 V

G = -n*F*Eo = -1*F*0.771 = - 0.771*F

from the data table we have:

Fe2+ (aq) + 2e- —> Fe (s) … eqn 2

Eo = -0.440 V

G = -n*F*Eo = -2*F*(-0.440) = 0.880*F

we need to find for :

Fe3+ (aq) + 3e- = Fe(s) …eqn 3

Eo= ?

G = -n*F*Eo = -3*F*Eo

eqn 3 can be obtained by adding eqn1 and eqn 2,

so,

G3 = G1 + G 2

-3*F*Eo = - 0.771*F + 0.880*F

-3*F*Eo = 0.109*F

-3*Eo = 0.109

Eo = - 0.036 V

Answer: -0.036 V

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