Question

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?

Answer 1

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

Step-by-step 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.

Answer 2

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