question archive a) Calculate the standard state potential and the equilibrium constant for the following unbalanced reaction
a) Calculate the standard state potential and the equilibrium constant for the following unbalanced reaction
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a) Calculate the standard state potential and the equilibrium constant for the following unbalanced reaction.
(Eo (Fe3+/Fe2+) = 0.771V, Eo (I3-/3I-) = 0.536V)
Fe3+ + I- -----> Fe2+ + I3-
b) Calculate the potential for the above when [Fe3+] = 0.025 M, [Fe2+] = 0.015 M, [I-] = 0.020 M and [I3-] = 0.025 M
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a. standard state potential = 0.235 V
equilibrium constant = 2.81 x 103
b. potential = 0.0270 V
Step-by-step explanation
a. First, balance the redox reaction.
1. Separate the half-reactions by separating similar species as shown below.
| Fe3+ -> Fe2+ | I- -> I3- |
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Since the elements are already balanced, balance the charge by adding 1 electron in the reactant side: Fe3+ + e- -> Fe2+ |
To balance I, add 3 as the coefficient to I-: 3I- -> I3-
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Fe3+ + e- -> Fe2+
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Then, balance the charges by adding 2 electrons in the product side: 3I- -> I3- + 2e-
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Multiply the whole equation by 2 so that the number of electrons in this half-reaction will be equal to the number of electrons of the the other half-reaction: 3I- -> I3- + 2e-: 2(Fe3+ + e- -> Fe2+) = 2Fe3+ + 2e- -> 2Fe2+
***Note that this is the reduction half reaction since electrons are gained in this reaction |
3I- -> I3- + 2e-
***Note that this is the oxidation half reaction since electrons are released in this reaction |
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Add the two half-reactions: 2Fe3+ + 2e- -> 2Fe2+ + 3I- -> I3- + 2e- = 2Fe3+ + 2e- + 3I- -> 2Fe2+ + I3- + 2 e- Cancel out 2e- from the reactant and product side. 2Fe3+ + 3I- -> 2Fe2+ + I3- |
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Calculate the standard state potential using the formula:
Eoreaction = Eoreduction - Eooxidation
Where Eoreaction = standard state potential of the reaction, Eoreduction = standard state potential of the reduction half-reaction, and Eooxidation = standard state potential of the oxidation half-reaction.
Eoreaction = Eoreduction - Eooxidation = 0.771V - 0.536V = 0.235 V
Then calculate the equilibrium constant using the formula:
Ereactiono?=n0.0592?lnK
Where Eoreaction = standard state potential of the reaction, n = number of electrons and K = equilibrium constant.
Here, the standard state potential of the reaction is calculated above, the number of electrons, n is equal to 2 as shown in the balanced equation above while K is the unknown which is the equilibrium constant.
Ereactiono?=n0.0592?lnK
0.235=20.0592?lnK
0.235×0.05922?=lnK
e7.939=elnK
e7.939=K
K=2.81×103
b) Now, for non-standard state like this example, use the formula for non-standard state using the formula below:
Ereaction?=Ereactiono?−n0.0592?lnK
Where Ereaction = non-standard state potential of the reaction, Eoreaction = standard state potential of the reaction, n = number of electrons and K = equilibrium constant.
Here, we already calculated Eoreaction above and the number of electrons is 2. Hence, we just need to determine the expression for K. Note that K, the equilibrium constant is equal to the ratio of the product of the concentration of the products and the product of the concentrations of the reactants with exponent equal to their respective coefficients in the balanced chemical reaction (shown above), Hence,
K=[Fe3+]2[I−]3[Fe2+]2[I3−?]?
Input this to the equation above and calculate for Ereaction as shown below:
Ereaction?=Ereactiono?−n0.0592?ln[Fe3+]2[I−]3[Fe2+]2[I3−?]?
Ereaction?=0.235V−20.0592?ln(0.025M)2(0.020M)3(0.015M)2(0.025M)?=0.0270V
