Question & Answer: Standard reduction potentials Use the table of standard reduction potentials given above to…..

Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e-→Ag(s) 0.80 Cu2+ (aq) + 2e-→Cu(s)| 0.34 Sn4+(aq) 4e-+Sn(s) 0.15 0 Ni2+(aq) + 2e-→Ni(s) 0.26 Fe2+(aq) 2e-Fe(s)-0.45 Zn2+ (aq) + 2e-Zn(s)-076 A12+(aq) +3e--AI(s) | -1.66 Mg2+(aq) + 2e-→Mg(s)| -2.37 2H+ (aq) + 2e-→H2 (g) Part A Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ° C) for the following reaction: Fe(s) Ni2+ (aq) →Fe2+ (aq) Ni(s) Express your answer numerically.

Standard reduction potentials Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 degree C) for the following reaction: Fe(s) + Ni^2+ (aq) rightarrow Fe^2+ (aq) + Ni(s) Express your answer numerically.

Expert Answer

Answer

First, relate Ecell to Keq:

dG = -n*F*Ecell

dG = -R*T*ln(Keq)

where, n is the number of moles, F is farady constant 96500, Ecell, the Eº calcualted , R ideal gas constant, T temperature

solve for Keq

Keq = exp(nF*Ecell/(RT))

We need Eºcell

Eºcell = Ecathode – Eanode = -0.26 – (-0.45) = 0.19 V

now, n = 2 moles of e-

substitute all data:

Keq = exp(nF*Ecell/(RT))

Keq = exp(2*96500*0.19/(8.314*298))

Keq = 2678535.54417

Keq = 2.68*10^6

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