Question

Use the Activity Series (Silberberg) to predict whether each of the following reactions will occur spontaneously (yes or no):
a) Mn²⁺ (aq) + Co (s) --> Mn (s) + Co²⁺ (aq)
b) 3 Zn (s) + 2 Cr³⁺ (aq) --> 3 Zn²⁺ (aq) + 2 Cr (s)
c) Ba²⁺ (aq) +2 Li (s) --> Ba (s) + 2 Li⁺ (aq)
d) Hg²⁺ (aq) + Cd (s) --> Hg (l) + Cd²⁺ (aq)
e) Ca (s) + 2 K⁺ (aq) --> Ca²⁺ (aq) + 2 K (s)
f) Ni²⁺ (aq) + Mg (s) --> Ni (s) + Mg²⁺ (aq)

For each of these reactions, calculate the overall cell voltage, (using Append D Silberberg) and compare this quantitative prediction with the qualitative Activity Series predictions (spontaneous or not) listed above.

Answer 1

a) Mn²⁺ (aq) + Co (s) --> Mn (s) + Co²⁺ (aq)

From the activity series, Mn lies above Co; which means Mn is more reactive than Co and should undergo oxidation. So, this reaction is not spontaneous.

Oxidation:   Co (s) --> Co²⁺ (aq) E^o = 0.28 V

Reduction:Mn²⁺ (aq) --> Mn (s) E^o = -1.18 V

E^o_cell = 0.28 - 1.18 = -0.9 V

Since cell potential is negative, it is not a spontaneous reaction.

b) 3 Zn (s) + 2 Cr³⁺ (aq) --> 3 Zn²⁺ (aq) + 2 Cr (s)

From the activity series, Zn lies above Cr; which means Zn is more reactive than Cr and should undergo oxidation. So, this reaction is spontaneous.

Oxidation:  3 Zn (s) --> 3 Zn²⁺ (aq) + 6 e^-    3 x ( E^o = 0.76 )

Reduction: 2 Cr³⁺ (aq) + 6 e^- --> 2 Cr (s) 2 x ( E^o = -0.74)

E^o_cell = 3 x (0.76) + 2 x (-0.74) = 0.80 V

Since cell potential is positive, the reaction is spontaneous.

c) Ba²⁺ (aq) + 2 Li (s) --> Ba (s) + 2 Li⁺ (aq)

From the activity series, Li lies above Ba; which means Li is more reactive than Ba and should undergo oxidation. So, this reaction is spontaneous.

Oxidation:  2 Li (s) --> 2 Li⁺ (aq) + 2 e^-    2x ( E^o = 3.04 )

Reduction: Ba²⁺ (aq) + 2 e^- --> Ba (s)   E^o =-2.91 V

E^o_cell = 2(3.04) +(-2.91) = 3.17

Since the cell potential is positive, the reaction is spontaneous.

d) Hg²⁺ (aq) + Cd (s) --> Hg (l) + Cd²⁺ (aq)

From the activity series, Cd lies above Hg; which means Cd is more reactive than Hg and should undergo oxidation. So, this reaction is spontaneous.

Oxidation: Cd (s) --> Cd²⁺ (aq) + 2 e^- E^o = 0.40

Reduction: Hg²⁺ (aq) + 2 e^- --> Hg (l) E^o = 0.85 V

E^o_cell = 0.40 + 0.85 = 1.25 V

Since the cell potential is positive, the reaction is spontaneous.

e) Ca (s) + 2 K⁺ (aq) --> Ca²⁺ (aq) + 2 K (s)

From the activity series, K lies above Ca; which means K is more reactive than Ca and should undergo oxidation. So, this reaction is not spontaneous.

Oxidation: Ca (s) --> Ca²⁺ (aq) + 2 e^-         E^o = 2.87 V

Reduction: 2 K⁺ (aq) + 2 e^- --> 2 K (s)    2 x ( E^o = -2.93 V )

E^o_cell = 2.87 + 2 x (- 2.93) = -2.99 V

Since the cell potential is negative, the reaction is not spontaneous.

f) Ni²⁺ (aq) + Mg (s) --> Ni (s) + Mg²⁺ (aq)

From the activity series, Mg lies above Ni; which means Mg is more reactive than Ni and should undergo oxidation. So, this reaction is spontaneous.

Oxidation: Mg (s) --> Mg²⁺ (aq) + 2 e^- E^o = 2.37 V

Reduction: Ni²⁺ (aq) + 2 e^- --> Ni (s)    E^o = -0.26 V

E^o_cell = 2.37 - 0.26 = 2.11 V

Since the cell potential is positive, the reaction is spontaneous.