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
2H+(aq)+2e−→H2(g) 0
Ni2+(aq)+2e−→Ni(s) −0.26
Fe2+(aq)+2e−→Fe(s) −0.45
Zn2+(aq)+2e−→Zn(s) −0.76
Al3+(aq)+3e−→Al(s) −1.66
Mg2+(aq)+2e−→Mg(s) −2.37
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.

Here magnesium (Fe) undergoes oxidation by loss of electrons, thus act as anode. silver (Ni) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

[tex]E^0_{[Fe^{2+}/Fe]}=-0.45V[/tex]

[tex]E^0_{[Ni^{2+}/Ni]}=-0.26V[/tex]

[tex]E^0=E^0_{[Ni^{2+}/Ni]}-E^0_{[Fe^{2+}/Fe]}[/tex]

[tex]E^0=-0.26V-(-0.45V)=0.19V[/tex]

Now we have to calculate the Gibbs free energy.

Formula used :

[tex]\Delta G^o=-nFE^o[/tex]

where,

[tex]\Delta G^o[/tex] = Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

[tex]E^o[/tex] = standard e.m.f of cell = 0.19 V

Now put all the given values in this formula, we get the Gibbs free energy.

[tex]\Delta G^o=-(2\times 96500\times 0.19)=-36670J/mole[/tex]

Now we have to calculate the value of equilibrium constant.

Formula used :

[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = [tex]25^oC=273+25=298K[/tex]

[tex]K_c[/tex] = equilibrium constant = ?

Now put all the given values in this formula, we get the value of [tex]K_c[/tex]

[tex]-36670J/mole=-2.303\times (8.314 J/K/mole)\times (298K)\times \log K_c[/tex]

[tex]K_c=2.667\times 10^{6}[/tex]

Therefore, the value of [tex]K_c[/tex] at 298 K is, [tex]2.667\times 10^{6}[/tex]

pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-01-24T14:34:06+01:00

The equilibrium constant as required is 2.67 × 10^6.

We already know that;

ΔG∘ = -nFE∘cell

ΔG∘ = change in free energy
n = number of moles of electrons transferred
F = Faraday constant
E∘cell = standard cell potential

We now have all the required redox potentials in the list as stated in the question. Hence;

E∘cell = (−0.26 V) - (−0.45 V)

E∘cell = 0.19 V

Then;

ΔG∘ = -(2 × 96500 × 0.19 V)

ΔG∘ = -36.67 kJ

Also;

ΔG∘ = -RTlnK

lnK = ΔG∘/-RT

lnK = -36.67 × 10^3 J/-(8.314 J/K/mol × 298 K)

lnK = 14.8

e^lnK = e^14.8

K = 2.67 × 10^6

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