Analyze and solve this partially completed galvanic cell puzzle. There are 4 ele
ID: 909444 • Letter: A
Question
Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alphabet, A through D. The values in the partially completed grid are measured cell potentials for a cell consisting of electrode #1 and electrode #2. You may assume that each galvanic cell was properly constructed with the appropriate metals and solutions and that all the measured values in the grid are accurate.
electrode #1 ?
C
B
D
A
electrode #2 ?
Ecell(volts)
Ecell(volts)
Ecell(volts)
Ecell(volts)
C
0
0.91
0.62
0.01
B
0.91
0
1.53
0.90
D
0.62
1.53
0
0 volts
0.01 volts
0.28 volts
0.61 volts
0.63 volts
0.89 volts
0.90 volts
0.91 volts
0.92 volts
1.52 volts
1.54 volts
1.81 volts
2.43 volts
In this experiment you worked with 4 electrodes (zinc, copper, tin and silver/silver chloride) and constructed a table of reduction potentials with these 4 electrodes. Use that information to arrange these 4 electrodes in order from best reducing agent to poorest reducing agent.
copper
tin
zinc
silver/silver chloride
Select the true statements from the list below. One or more are correct. You will receive negative points for incorrect answers.
tin metal will reduce Cu2+ to copper metal
The cell potential for a cell consisting of tin metal immersed in a tin(II) solution and the standard hydrogen electrode is not known because it was not measured in this experiment.
Zn2+ is the best oxidizing agent studied in this experiment
Zn2+ is the best reducing agent studied in this experiment
copper metal will not reduce H+ to hydrogen gas
Hydrogen gas is a better reducing agent than tin metal.
Cu2+ is the best oxidizing agent studied in this experiment.
zinc metal is the best reducing agent studied in this experiment.
electrode #1 ?
C
B
D
A
electrode #2 ?
Ecell(volts)
Ecell(volts)
Ecell(volts)
Ecell(volts)
C
0
0.91
0.62
0.01
B
0.91
0
1.53
0.90
D
0.62
1.53
0
Explanation / Answer
1) Given
E0B - E0A = 0.90 ......(1)
E0B - E0D = 1.53 .......(2)
E0A - E0D = ? ....(3)
So we can obtain (3) from 1 and 2 as
Eq(3) = Eq(2) - E1(1) = 0.63 V
2) the value of reduction potentials of given elements are
E0Zn+2/ Zn = -0.76 V
E0Cu+2/ Cu = + 0.34 V
E0Sn+2/ Sn =-0.14 V
E0AgCl/ Ag = +0.22 V
So the best to poorest reducing agent order will be
Zinc > Tin > Silver > copper
3) True statemenst are
tin metal will reduce Cu2+ to copper metal
copper metal will not reduce H+ to hydrogen gas
Cu2+ is the best oxidizing agent studied in this experiment.
zinc metal is the best reducing agent studied in this experiment.
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