Use the Activity Series (Silberberg) to predict whether each of the following re
ID: 866500 • Letter: U
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.
Explanation / Answer
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) Eo = 0.28 V
Reduction:Mn2+ (aq) --> Mn (s) Eo = -1.18 V
Eocell = 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 ( Eo = 0.76 )
Reduction: 2 Cr3+ (aq) + 6 e- --> 2 Cr (s) 2 x ( Eo = -0.74)
Eocell = 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 ( Eo = 3.04 )
Reduction: Ba2+ (aq) + 2 e- --> Ba (s) Eo =-2.91 V
Eocell = 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- Eo = 0.40
Reduction: Hg2+ (aq) + 2 e- --> Hg (l) Eo = 0.85 V
Eocell = 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- Eo = 2.87 V
Reduction: 2 K+ (aq) + 2 e- --> 2 K (s) 2 x ( Eo = -2.93 V )
Eocell = 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- Eo = 2.37 V
Reduction: Ni2+ (aq) + 2 e- --> Ni (s) Eo = -0.26 V
Eocell = 2.37 - 0.26 = 2.11 V
Since the cell potential is positive, the reaction is spontaneous.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.