we constructed galvanic cells with the following solutions (with a their respect
ID: 756589 • Letter: W
Question
we constructed galvanic cells with the following solutions (with a their respective metals used as cathodes): KNO3 was used to construct a salt bridge to all the following solutions: ZnSO4, CuSO4, FeSO4, NI(NO3)2, KI/I2. we then used a multimeter to test voltaages, this is what we came up with for results: Cell Anode Cathod Voltage Assuming that the reduction potential for the most active metal is 0 V Zn + Cu Zn Cu 1.01 Using observed voltages determine the reduction potentials for the other Zn+Fe Zn Fe .39 oxidized species investigated. Zn+NI Zn Ni .25 what was strongest oxidizing agent?/reducing agent? Zn+I2/I- Zn I2/I- 1.31 what is the highest cell voltage that could be used from the half Fe+Ni Ni Fe .16 cells used in this experiment? Fe + I2/I- Fe I2/I- .92 supposed taht Fe/Fe2+ was used as the reference reaction instead of the one that included the most active metal. what would Ni+I2/I- Ni I2/I- 1.03 the highest cell voltage that could be constructed from 1/2 cells be? determine value for delta G0 for the following systems using your value for E0 Zn + Fe2+ = Zn2+ + Fe Cu + Zn2+ = Cu2+ + Zn order the following in order of activity, w/ most active first: Cu, H, Fe, Ni, Zn, Na, Mg, AgExplanation / Answer
( **Note: There is some formatting error in this question. I have tried to correct it. PLease check if all the values are mentioned correctly.)
Cell Anode Cathode Voltage
Zn + Cu Zn Cu 1.01
Zn + Fe Zn Fe .39
Zn + I2/I- Zn I2/I- 1.31
Fe + Ni Ni Fe .16
Fe + I2/I- Fe I2/I- .92
Ni + I2/I- Ni I2/I- 1.03
Using the activity series, we know that Zn is the most active metal. Therefore we can assume reduction potential of Zn as 0 V .
Let's find reduction potential of other species.
Assuming we have used solutions of 1 M concentration, we have Ecell = E^0 cell
E^0 cell = E^0 cathode - E^0 anode ( We will be using this formula to calculate reduction potentials of other species)
First cell we have is ""Zn + Cu"" where Zn is anode and Cu is cathode
E^0 cell = E^0 (Cu) - E^0 (Zn)
1.01 = E^0 (cu) - 0
E^0 (Cu) = 1.01 V
Cell --> Zn + Fe , Here Zn is anode and Fe is cathode
E^0 Cell = E^0 (Fe) - E^0 ( Zn)
0.39 = E^0 (Fe) - 0
E^0 (Fe) = 0.39 V
Cell --> Zn + I2/I- Here Zn is anode and I2/i- is cathode
E^0 cell = E^0 (I2/I-) - E^0 (zn)
1.31 = E^0 (I2/I-) - 0
E^0 (I2/I-) = 1.31 V
Cell --> Fe + Ni Here, Ni is anode and Fe is cathode
E^0 cell = E^0 (Fe) - E^0 (Ni)
0.16 = 0.39 - E^0 (Ni) ---------( substituted E^(Fe) from previous calculation
E^0 (Ni) = -0.23 V
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Oxidizing agent is the substance which itself undergoes reduction.
Therefore in order to be strongest oxidizing agent, the substance has to undergo reduction easily.
The ease with which a species can be reduced is indicated by reduction potential
Higher the value of reduction potential, easier for the species to undergo reduction.
Therefore, Strongest oxidizing agent will have highest reduction potential , from the above calculations , we know E^0 is highest for I2/I- half cell.
Strongest oxidizing agent is I2
Strongest reducing species would be the one which has lowest reduction potential.
From the calculated values, we know Ni has lowest value
Strongest reducing species is Ni
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In order to have highest cell voltage, we would select our cathode species which has highest value of reduction potential.
From the calculated values, it is I2/I- ( 1.31 V)
Anode species would be the one which has lowest reduction potential. From the calculated values, it is Ni ( E^0 = -0.23 V)
A cell constructed using I2/I- as cathode and Ni as anode would have highest cell voltage.
It can be calculated as follows
E^0 cell = E^0 (I2/I-) - E^0 (Ni)
E^0 cell = 1.31 V - ( -0.23V)
E^0 cell = 1.54 V
The highest cell voltage that can be generated is 1.54 V
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If Fe was used as the reference electrode, then we have to assume E^0 (Fe) = 0.
Let's find potential of other species using Fe as a reference
Cell --> Zn + Fe , Here we have Zn as anode and Fe as cathode
E^0 cell = E^0 (Fe) - E^0 (Zn)
0.39 = 0 - E^0 (Zn)
E^0 (Zn) = -0.39 V
Cell --> Fe + Ni , Here we have Ni as anode and Fe as cathode
E^0 cell = E^0 ( Fe) - E^0 (Ni)
0.16 = 0 - E^0 ( Ni)
E^0 ( Ni) = -0.16 V
Cell --> Fe + I2/I- , Here we have Fe as anode and I2/I- as cathode
E^0 cell = E^0 (I2/I-) - E^0 (Fe)
0.92 = E^0 ( I2/I-) - 0
E^0 (I2/I-) = 0.92 V
To have highest cell voltage, we need cathode with highest E^0 value and anode with lowest E^0 value
From the values, calculated above , we can select cathode as I2/I- and anode as Zn
E^0 cell = E^0 ( I2/I- ) - E^0 (Zn)
E^0 cell = 0.92 - ( -0.39)
E^0 cell = 1.31 V
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The given reaction is Zn + Fe^2+ <------> Zn^2+ + Fe
Let's write half reactions
Zn -----> Zn^2+ + 2e- ..........Oxidation
Fe^2+ + 2e- ------> Fe ..........Reduction
Reduction occurs at cathode, therefore we have cathode as Fe and anode as Zn
E^0 cell value for this system is 0.39 V
E^0 and Delta G^0 are related by following equation
Delta G^0 = - n* F* E^0 cell
For the above system, n = number of electrons transferred which is 2 e-
F is Faraday's constant which is 96500
E^0 cell = 0.39 V
Substituting above values, we get
Delta G^0 = - 2 * 96500 C * 0.39 V
Delta G^0 = - 75270 C. V
The unit that we have above is coulomb.volt which can be related to unit of energy ad follows
1 C.V = 1 J
therefore 75270 C.V = 75270 J
Delta G^0 = - 75270 J
Delta G^0 = - 75.27 kJ
The next reaction is Cu + Zn2+ <---------> Cu2+ + Zn
Cu -----> Cu^2+ + 2e- .................... Oxidation
Zn^2+ + 2e- -------> Zn ................Reduction
Here, Zn is cathode and Cu is anode
This is set up is exactly opposite to that we have in our experiment.
Therefore E^cell values would also get reversed in this case.
E^0 cell = -1.01 ( for a system of cathode = Cu and anode = Zn we had E^0 cell = 1.01 . This is revese system , so we reverse the potential as well)
Using the same equation we used earlier, we get
Delta G^0 = - ( -1.01 ) * 96500 * 2
Delta G^0 = 194930 J
Delta G^0 = 194.9 ~ 195 kJ
Delta G^0 = 195 kJ
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The given elements are Cu, H, Fe, Ni, Zn, Na, Mg, Ag
Using the standard reduction potential data table we can arrange the given substances in following order
Element having lowest value of reduction potential are most active
Na > Mg > Zn > Fe > Ni > H > Cu > Ag
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