Standard reduction half-cell potentials at 25C. THIS CHART IS FOR BOTH QUESTIONS

Question

Standard reduction half-cell potentials at 25C. THIS CHART IS FOR BOTH QUESTIONS BELOW:

QUESTION 1) Use tabulated standard electrode potentials to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 C. (The equation is balanced.)

2Fe3+(aq)+3Zn(s)2Fe(s)+3Zn2+(aq)

QUESTION 2) Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+:

Ni(s)+2H+(aq)Ni2+(aq)+H2(g)

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

2Fe3+(aq)+3Zn(s)2Fe(s)+3Zn2+(aq)

Fe3+(aq)+3eFe(s) 0.036

Zn2+(aq)+2eZn(s) 0.76

Zn oxidizes:

Fe3+(aq)+3eFe(s)

Zn(s)Zn2+(aq)+2e

E = 0.76 - 0.036 = 0.724 V

2)

Ni(s)+2H+(aq)Ni2+(aq)+H2(g)

Ni2+(aq)+2eNi(s) 0.23

2H+(aq)+2eH2(g)   0.00

Nickel will oxidize

Ni(s) Ni2+(aq)+2e +0.23

2H+(aq)+2eH2(g)   0.00

Ecell = 0.23

G = -nFEcell

G = -RT*lnK

Then

RT*lnK = nFEcell

8.314*298*lnK = 2*96500*0.23

lnK = 17.916

K = exp(17.916) = 6.036*10^7

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