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?

Answers

Answer 1

Answer: 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 2

Answer:

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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