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Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions. Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) S (s) (nonspontaneous)? Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq) (nonspontaneous)? 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq) (spontaneous?) Are these correct?

Respuesta :

PBCHEM
A electrochemical reaction is said to be spontaneous, if [tex] E^{0} cell is positive. [/tex] 

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    [tex] S^{2-}/S // Ni^{2+}/Ni [/tex]

Hence, [tex] E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-} [/tex]
we know that, [tex]{E^{0} Ni^{2+}/Ni = -0.25 v [/tex]
and [tex]{E^{0} S/ S^{2-} = -0.47 v [/tex]

Therefore, [tex]E^{0} cell [/tex] = - 0.25 - (-0.47) = 0.22 v

Since,  [tex]E^{0} cell [/tex] is positive, hence cell reaction is spontaneous
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Answer 2: 
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
    [tex]H_{2} / H^{+} // Pb^{2+} /Pb[/tex]

Hence, [tex] E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+} [/tex]
we know that, [tex]{E^{0} Pb^{2+}/Pb = -0.126 v [/tex]
and [tex]{E^{0} H_{2}/ H^{+} = -0 v [/tex]

Therefore, [tex]E^{0} cell [/tex] = - 0.126 - 0 = -0.126 v

Since,  [tex]E^{0} cell [/tex] is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
    [tex] Cr/Cr^{2+} // Ag^{+}/Ag [/tex]

Hence, [tex] E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+} [/tex]
we know that, [tex]{E^{0} Ag^{+}/Ag = -0.22 v [/tex]
and [tex]{E^{0} Cr/ Cr^{2+} = -0.913 v [/tex]

Therefore, [tex]E^{0} cell [/tex] = - 0.22 - (-0.913) = 0.693 v

Since,  [tex]E^{0} cell [/tex] is positive, hence cell reaction is spontaneous

Answer: [tex]Ni^{2+}(aq)+S^{2-}(aq)\rightarrow Ni(s)+S(s)[/tex]  : non spontaneous

[tex]Pb^{2+}(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)[/tex]  : non spontaneous

[tex]2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)[/tex]  : spontaneous

Explanation:

a) [tex]Ni^{2+}(aq)+S^{2-}(aq)\rightarrow Ni(s)+S(s)[/tex]

Here S undergoes oxidation by loss of electrons, thus act as anode. Ni undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Ni^{2+}/Ni]}=-0.25V[/tex]

[tex]E^0_{[S^{2-}/S]}=0.407VV[/tex]

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

[tex]E^0=-0.25-(0.407V)-0.657V[/tex]

As value of [tex]E^0[/tex] is negative, the reaction is non spontaneous.

b) [tex]Pb^{2+}(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)[/tex]

Here Hydrogen undergoes oxidation by loss of electrons, thus act as anode. Pb undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Pb^{2+}/Pb]}=-0.13[/tex]

[tex]E^0_{[H^{+}/H_2]}=0V[/tex]

[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[H^{+}/H_2]}[/tex]

[tex]E^0=-0.13-(0V)=-0.13V[/tex]

As value of [tex]E^0[/tex] is negative, the reaction is non spontaneous.

c) [tex]2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)[/tex]

Here Cr undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Ag^{+}/Ag]}=+0.80V[/tex]

[tex]E^0_{[Cr^{2+}/Cr]}=-0.913V[/tex]

[tex]E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Cr^{2+}/Cr]}[/tex]

[tex]E^0=+0.80-(-0.913V)=1.713V[/tex]

As value of [tex]E^0[/tex] is positive, the reaction is spontaneous.

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