Question

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→fe(s) −0.45 ag (aq) e−→ag(s) 0.80 cr3 (aq) e−→cr2 (aq) −0.50 fe3 (aq) 3e−→fe2 (aq) 0.77 cr3 (aq) 3e−→cr(s) −0.73 cu (aq) e−→cu(s) 0.52 zn2 (aq) 2e−→zn(s) −0.76 cu2 (aq) 2e−→cu(s) 0.34 mn2 (aq) 2e−→mn(s) −1.18 2h (aq) 2e−→h2(g) 0.00 al3 (aq) 3e−→al(s) −1.66 fe3 (aq) 3e−→fe(s) −0.036 mg2 (aq) 2e−→mg(s) −2.37 pb2 (aq) 2e−→pb(s) −0.13 na (aq) e−→na(s) −2.71 sn2 (aq) 2e−→sn(s) −0.14 ca2 (aq) 2e−→ca(s) −2.76 ni2 (aq) 2e−→ni(s) −0.23 ba2 (aq) 2e−→ba(s) −2.90 co2 (aq) 2e−→co(s) −0.28 k (aq) e−→k(s) −2.92 cd2 (aq) 2e−→cd(s) −0.40 li (aq) e−→li(s) −3.04 you may want to reference (pages 749 - 752) section 18.5 while completing this problem. part a use the tabulated electrode potentials to calculate k for the oxidation of zinc by h : zn(s) 2h (aq)→zn2 (aq) h2(g)

Answer 1

K = 5.0 × 10^25

Part (a). Calculate E° for the reaction

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

Part (b). Calculate K for the reaction

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25