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Black_prince [1.1K]
3 years ago
7

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?
Chemistry
2 answers:
sammy [17]3 years ago
8 0
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

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

The cell representation of above reaction is given by;
    </span>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: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
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: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
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
solong [7]3 years ago
8 0

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.

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levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

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Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

           [Ca^{2+}] = 0.00625 M

Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

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