other in another container, some means must be provided to maintain charge neutr
ID: 513666 • Letter: O
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other in another container, some means must be provided to maintain charge neutrality of the two containers. If we remove electrons from one container that container win become positively charged. This charging will have an adverse effect on the cell and no potential will be observed. In order to prevent this a "salt bridge" is used to connect one container to the other. The salt bridge usually contains some simple salt that is inert under the condition of the reactions. For example, KNOB is commonly used because neither K- nor NO3- will under go a redox reaction in a typical cell. As electrons are removed from one container, NO3- ions flow out of the salt bridge into the container and compensate for the charge removed. From an electronics point of view we must have a complete path for current flow One that originates at one electrode and ultimately returns to that electrode. The bridge serves this purpose. Although electrons do not flow in the bridge, charges in the form of cations and ions do flow. Pre-Laboratory Assignment 1. Using the data in Table Icalculate the standard potentials of the following cells a. Cu /Cu and Zn? Zn b. Cua/Cuand Pba/Pb Cu /Cu and Ag /Ag d. Cu "/Cu and Cd? /Cd 2 Using the Nernst Equation calculate the potential of the concentration cell made wit 0.1 MZna/Znand 0.01 M Zn2 /Zn. 9. Using data in Table I answer the following: a. Will Zn metal spontaneously plate ontoacopper rod when the rod is placed in solution of Zn b. Will the reverse reaction occur? Electrochemical Cells and Redox ReactionsExplanation / Answer
Answer for (1a):
Cu2+/Cu and Zn2+/Zn cell is conventionally written as
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Net reaction of the above cell:
Zn(s) + Cu2+ Zn2+ + Cu(s)
Half reactions
Zn(s) Zn2+ + 2e– ..................oxidation E0 = 0.76 V (from given table)
Cu2+ + 2e– Cu(s) .................reduction E0 = +0.34 V (from given table)
Standard potential for the reaction, E0 = E0 reduction - E0 oxidation
Therefore E0 = 0.34 - (0.76)
E0 = + 1.10 V
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Answer for (1b):
Cu2+/Cu and Pb2+/Pb cell is conventionally written as
Pb(s) | Pb2+(aq) || Cu2+(aq) | Cu(s)
Net reaction of the above cell:
Pb(s) + Cu2+ Pb2+ + Cu(s)
Half reactions
Pb2+ + 2e– Pb(s) ..................reduction E0 = 0.13 V (from given table). Among the electrodes lead reduction is negative and is taken as oxidative reactoin.
Cu2+ + 2e– Cu(s) .................reduction E0 = +0.34 V (from given table)
Standard potential for the reaction, E0 = E0 reductionhigher - E0 reduction lower
Therefore E0 = 0.34 - (0.13)
E0 = + 0.47 V
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Answer for (1c):
Cu2+/Cu and Ag+/Ag cell is conventionally written as
2Ag(s) | 2Ag+(aq) || Cu2+(aq) | Cu(s)
Net reaction of the above cell:
2Ag(s) + Cu2+ 2Ag+ + Cu(s)
Half reactions
2Ag+ + 2e– 2Ag(s) ..................reduction E0 = 1.60 V (from given table). Among the electrodes silver reduction value is higher than Cu2+ reduction and is taken as reductive reactoin.
Cu2+ + 2e– Cu(s) .................reduction E0 = +0.34 V (from given table)
Standard potential for the reaction, E0 = E0 reductionhigher - E0 reduction lower
Therefore E0 = 1.6 - 0.34
E0 = + 1.26 V
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Answer for (1d):
Cu2+/Cu and Cd2+/Cd cell is conventionally written as
Cd(s) | Cd2+(aq) || Cu2+(aq) | Cu(s)
Net reaction of the above cell:
Cd(s) + Cu2+ Cd2+ + Cu(s)
Half reactions
Cd2+ + 2e– Cd(s) ..................reduction E0 = -0.40 V (from given table). Among the electrodes cadmium reduction value is lower than Cu2+ reduction and is taken as oxidative reactoin.
Cu2+ + 2e– Cu(s) .................reduction E0 = +0.34 V (from given table)
Standard potential for the reaction, E0 = E0 reduction - E0oxidation
Therefore E0 = 0.34 -(- 0.40)
E0 = + 0.74 V
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Answer for (2):
Concentration cell: 0.1M Zn2+/Zn and 0.01M Zn2+/Zn and the cell is conventionally written as
0.1M Zn2+| Zn || 0.01M Zn2+|Zn =
Nernst equation, Ecell = E°cell – (2.303 RT / n F) log Q
Under standard conditions the equation becomes
Ecell = E°cell – (0.0592 / n )log Q
where
E°cell = the standard cell potential
n = the number of moles of electrons transferred
Q = [product ion] / [reactant ion]
Accordingly E°cell = -0.76V - (0.0592 / 2 )log 0.01/0.1
Ecell = -0.76 – (0.0592 / 2 )log 0.1
Ecell = -0.76 – (0.0296) log 0.1
= -0.76 - (0.0296) (-1)
= -0.76 - (-0.0296)
Ecell = -0.7304 V
Answer for 3a = no and for 3b = yes=
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