b7laizbsoccergu
Answered

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?

Verified Answer :

PBCHEM

A electrochemical reaction is said to be spontaneous, if \(E^{0} cell is positive.\) 

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

The cell representation of above reaction is given by;
    \(S^{2-}/S // Ni^{2+}/Ni\)

Hence, \(E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-}\)
we know that, \({E^{0} Ni^{2+}/Ni = -0.25 v\)
and \({E^{0} S/ S^{2-} = -0.47 v\)

Therefore, \(E^{0} cell\) = - 0.25 - (-0.47) = 0.22 v

Since,  \(E^{0} cell\) is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
    \(H_{2} / H^{+} // Pb^{2+} /Pb\)

Hence, \(E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}\)
we know that, \({E^{0} Pb^{2+}/Pb = -0.126 v\)
and \({E^{0} H_{2}/ H^{+} = -0 v\)

Therefore, \(E^{0} cell\) = - 0.126 - 0 = -0.126 v

Since,  \(E^{0} cell\) 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;
    \(Cr/Cr^{2+} // Ag^{+}/Ag\)

Hence, \(E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}\)
we know that, \({E^{0} Ag^{+}/Ag = -0.22 v\)
and \({E^{0} Cr/ Cr^{2+} = -0.913 v\)

Therefore, \(E^{0} cell\) = - 0.22 - (-0.913) = 0.693 v

Since,  \(E^{0} cell\) is positive, hence cell reaction is spontaneous

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

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

\(2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)\)  : spontaneous

Explanation:

a) \(Ni^{2+}(aq)+S^{2-}(aq)\rightarrow Ni(s)+S(s)\)

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

\(E^0=E^0_{cathode}- E^0_{anode}\)

Where both \(E^0\) are standard reduction potentials.

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

\(E^0_{[S^{2-}/S]}=0.407VV\)

\(E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[S^{2-}/S]}\)

\(E^0=-0.25-(0.407V)-0.657V\)

As value of \(E^0\) is negative, the reaction is non spontaneous.

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

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

\(E^0=E^0_{cathode}- E^0_{anode}\)

Where both \(E^0\) are standard reduction potentials.

\(E^0_{[Pb^{2+}/Pb]}=-0.13\)

\(E^0_{[H^{+}/H_2]}=0V\)

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

\(E^0=-0.13-(0V)=-0.13V\)

As value of \(E^0\) is negative, the reaction is non spontaneous.

c) \(2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)\)

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

\(E^0=E^0_{cathode}- E^0_{anode}\)

Where both \(E^0\) are standard reduction potentials.

\(E^0_{[Ag^{+}/Ag]}=+0.80V\)

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

\(E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Cr^{2+}/Cr]}\)

\(E^0=+0.80-(-0.913V)=1.713V\)

As value of \(E^0\) is positive, the reaction is spontaneous.

Related Questions