A.) Use tabulated electrode potentials to calculate G rxn for each reaction at 2

Question

A.) Use tabulated electrode potentials to calculate Grxn for each reaction at 25C. O2(g)+2H2O(l)+2Cu(s)4OH(aq)+2Cu2+(aq).   Br2(l)+2I(aq)2Br(aq)+I2(s) Two significant figures and in kJ.

B.) Use the tabulated electrode potentials to calculate K for the oxidation of tin by H+:

Sn(s)+2H+(aq)Sn2+(aq)+H2(g)

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

A. G(rxn) = –nFE(cell) { n = 2 , f (faraday constant) = 96500}

E(cell) = Ecathode - E anode

O2(g)+2H2O(l)+2Cu(s)4OH(aq)+2Cu2+(aq).

G(rxn) = –nFE(cell)

E(cell) = Ecathode - E anode

G(rxn) =  -20 kJ

Br2(l)+2I(aq)2Br(aq)+I2(s) {n = 2 number of electron transfer}

G(rxn) =  -110 kJ

B. G(rxn) = –nFE(cell)   { n = 2, f = 96500 }

E(cell) = (0.00 - (-0.14) ) = 0.14 V

G(rxn) = -2 x 96500 x 0.14

G(rxn) = -27020 J/mol = -27.020 KJ/mol

G(rxn) = -2.303 RT log K { T 298 K , R = 8.314 }

-27020 J/mol = -2.303 x 8.314 x 298 x log K

log K = 4.73

K = 104.73 = 54386.63

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