A. A voltaic cell is constructed based on the reaction of Ag(CN) 2 - ( aq ) with
ID: 1018701 • Letter: A
Question
A. A voltaic cell is constructed based on the reaction of Ag(CN)2-(aq) with Cr(s) producing Ag(s) and Cr3+(s). Identify the correct cell diagram.
a.
Ag(CN)2-(aq) | Cr(s) || Ag(s) | Cr3+(aq)
d.
Ag(CN)2-(aq) | Ag(s) || Cr(s) | Cr3+(aq)
b.
Ag(s) | Cr3+(aq) || Ag(CN)2-(aq) | Cr(s)
e.
Cr(s) | Cr3+(aq) || Ag(CN)2-(aq) | Ag(s)
c.
Cr(s) | Cr3+(aq) || Ag(s) | Ag(CN)2-(aq)
B. Use the table of standard reduction potentials below to identify the metal or metal ion that is the strongest oxidizing agent.
Half-reaction
Pb4+ + 2 e- ® Pb2+
+1.80
Au3+ + 3 e- ® Au
+1.50
Fe3+ + 3 e- ® Fe
+0.771
I2 + 2 e- ® 2 I-
+0.535
Pb2+ + 2 e- ® Pb
-0.124
Al3+ + 3 e- ® Al
-1.66
Mg2+ + 2 e- ® Mg
-2.37
K+ + e- ® K
-2.93
a.
Pb4+
d.
K
b.
Pb2+
e.
Al
c.
K+
C. Silver tarnish (Ag2S) can be removed by immersing silverware in a hot solution of baking soda (NaHCO3) in a pan lined with aluminum foil; however, foul-smelling hydrogen sulfide gas (H2S) is produced. Which of the following statements is correct?
Silver/silver(I)
+0.799
Sulfur/hydrogen sulfide
+0.141
Aluminum/aluminum(III)
-1.662
Sodium/sodium(I)
-2.713
a.
Aluminum ions react with S2-, form aluminum sulfide, and gaseous carbon dioxide is released.
b.
Silver ions in the presence of the baking soda (NaHCO3) oxidize sulfide to elemental sulfur that attacks the aluminum foil, which produces aluminum sulfide.
c.
The aluminum acts as a reducing agent for the silver(I) in silver sulfide; then bicarbonate ion protonates the sulfide ion that is released.
d.
Aluminum is plated onto the silver surface, making it shiny again, and then the reaction of bicarbonate with aluminum oxide releases CO2.
e.
Silver in Ag2S reduces the aluminum, becomes metallic silver in the process, and releases hydrogen sulfide, H2S.
D. Using the following data, determine the standard cell potential for the electrochemical cell constructed using the following reaction: Zn(s) + Pb2+(aq) ® Zn2+(aq) + Pb(s).
Half-reaction Standard reduction potential
Zn2+(aq) + 2 e- ® Zn(s) -0.763
Pb2+(aq) + 2 e- ® Pb(s) -0.126
a.
+0.637 V
d.
-0.889 V
b.
-0.637 V
e.
+0.889 V
c.
+1.274 V
E. Based on the information in the table of standard reduction potentials below, what is the standard cell potential for an electrochemical cell that has iron (Fe) and magnesium (Mg) electrodes immersed in 1M Fe3+ and Mg2+ solutions? Also, identify the cathode.
Half-reaction
Pb4+ + 2 e- ® Pb2+
+1.80
Au3+ + 3 e- ® Au
+1.50
Fe3+ + 3 e- ® Fe
+0.771
I2 + 2 e- ® 2 I-
+0.535
Pb2+ + 2 e- ® Pb
-0.124
Al3+ + 3 e- ® Al
-1.66
Mg2+ + 2 e- ® Mg
-2.37
K+ + e- ® K
-2.93
a.
+1.60 V with Fe as the cathode
d.
-3.14 V with Mg as the cathode
b.
+3.14 V with Mg as the cathode
e.
+3.14 V with Fe as the cathode
c.
-3.14 V with Fe as the cathode
a.
Ag(CN)2-(aq) | Cr(s) || Ag(s) | Cr3+(aq)
d.
Ag(CN)2-(aq) | Ag(s) || Cr(s) | Cr3+(aq)
b.
Ag(s) | Cr3+(aq) || Ag(CN)2-(aq) | Cr(s)
e.
Cr(s) | Cr3+(aq) || Ag(CN)2-(aq) | Ag(s)
c.
Cr(s) | Cr3+(aq) || Ag(s) | Ag(CN)2-(aq)
Explanation / Answer
A) The anode half-cell is described first(Cr (s) | Cr3+(aq) ); the cathode half-cell(Ag(CN)2-(aq) | Ag(s)) follows. Within a given half-cell, the reactants are specified first and the products last(for example, in oxidation half cell representation should be (Ag(CN)2-(aq) | Ag(s)). The description of the oxidation reaction is first, and the reduction reaction is last; when you read it, your eyes move in the direction of electron flow.
so option (e) satisfies all the above rules for cell notation.
B)
metal/metal ion with HIGHEST NEGTIVE STANDARD REDUCTION POTENTIAL is strong REDUCING AGENT
metal/metal ion with HIGHEST POSITIVE STANDARD REDUCTION POTENTIAL is strong OXIDISING AGENT
So, option (a) is right as Pb4+ has more positive reduction potential,so it is strong oxidising agent for given options.
C) Aluminum has a lower ionization energy (energy required to remove electrons from an atom of the element) than silver.As a result, aluminum is oxidized (loses electrons and oxidation number increases), and silver is reduced (gains electrons and oxidation number is reduced).Depending on the amount of tarnish, the silver will be bright and the aluminum foil may be brown with tarnish (aluminum oxide), in a short while.The silver tarnish is "transferred" to the aluminum via reactions, which occur instantaneously, as follows:
3 Ag2S(s) + 2 Al(s)+ 3 H2O(l) -------> 6 Ag(s) + Al2O3(s) + 3 H2S(aq)
The baking soda (sodium bicarbonate) reacts with the (sulfur-smelling) H2S:
3 NaHCO3(aq) + 3 H2S(aq) --------> 3 NaHS(aq) + 3 H2O(l)+ 3 CO2(g)
by observing the above reactions, option (C) & option (E) are correct.
D) Eocell = Eoreduction + Eooxidation
first we have to find the reduction half reaction and oxidation half reaction by standard reduction potential values.
given values are standard reduction potentials.
it is easy to get Eoreduction (of reduction half reaction) is same as standard reduction potential for species of reduction reaction , i.e Eoreduction = -0.126 v
to get the value of Eooxidation of oxidation half reaction, we need to reverse the sign of given reduction potential for species of oxidation reaction, i.e Eooxidation = -( -0.763)v
finally, Eocell = Eoreduction + Eooxidation
Eocell = -0.126 +(-(-0.763))
Eocell = -0.126+0.763 = +0.637 V
so option (a) is correct
E) Inorder to know which is cathode and which is anode.....we have to observe the values of standard reduction potentials(SRP values) given in table.
from that SRP values,
metal/metal ion with LOW STANDARD REDUCTION POTENTIAL is strong REDUCING AGENT (under goes easy oxidation, here, Mg2+)
metal/metal ion with HIGH STANDARD REDUCTION POTENTIAL is strong OXIDISING AGENT(under goes easy reduction, here, Fe3+)
we know that OXIDATION is at ANODE
REDUCTION is at CATHODE
so ANODE is Mg , while CATHODE is Fe
now , Eocell = Eoreduction + Eooxidation
Eocell = +0.771 + (-(-2.37))
Eocell = 0.771+2.37 = +3.141 V
so option (e) is correct
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