Respuesta :
Answer:
The Emf of the given cell at [Fe2+] = 2.0 M and [Fe3+] = 1.9 M is 0.48 V
Explanation:
The half cell reaction can be written as :
Anode-Half (oxidation) :
[tex]Fe^{2+}\rightarrow Fe^{3+} + 1e^{-}[/tex] ......E = 0.77 V
(multiply this equation by 4 to balance the electrons)
Cathode-half (reduction)
[tex]4H^{+} +O_{2} + 4e^{-} \rightarrow 2H_{2}O[/tex]....E= 1.23 V
[tex]E^{0}_{cell} = E_{cathode} - E_{anode}[/tex]
[tex]E^{0}_{cell} = 1.23 - 0.77[/tex]
[tex]E^{0}_{cell} = 0.46 V[/tex]
According to Nernst Equation
[tex]E_{cell} = E^{0} - \frac{RT}{nF}lnQ[/tex]
[tex]E_{cell} = E^{0} - \frac{0.059}{n}logQ[/tex]
n = number of electron transferred in the cell reaction = 4
The balanced equation is :
[tex]4Fe^{2+} + 4H^{+} +O_{2} \rightarrow 4Fe^{3+} + 2H_{2}O[/tex]
[tex]E_{cell} = 0.46 - \frac{0.059}{4}logQ[/tex]
[tex]log Q = \frac{[Fe^{3+}]^{4}}{[Fe^{2+}]^{4}}[/tex]
[tex]log Q = \left ( \frac{[Fe^{3+}]}{[Fe^{2+}]} \right )^{4}[/tex]
[tex]log Q = \left ( \frac{1.9}{2.0} \right )^{4}[/tex]
Insert the value of log Q in Nernst Equation:
[tex]E_{cell} = 0.46 - \frac{0.059}{4} log\left ( \frac{1.9}{2.0} \right )^{4}[/tex]
(using :[tex]log^{a}b = a\log b[/tex]
)
[tex]E_{cell} = 0.46 - log\frac{1.9}{2.0}[/tex]
[tex]E_{cell} = 0.46 - log(0.95)[/tex]
[tex]E_{cell} = 0.46 -(-0.0227)[/tex]
[tex]E_{cell} = 0.482 V[/tex]
The cell EMF is obtained from Nernst equation as 0.451 V.
The reaction equation is given by;
4Fe2+ (aq) + O2(g) + 4H+ (aq)-------->4Fe3+(aq)+2H2O(l)
We have the following information;
EMF under standard conditions = 0.46 V
[Fe2+]= 2.0M
[Fe3+]= 1.9 M
From Nernst equation;
Ecell = E°cell - 0.0592/n logQ
Where;
E°cell = 0.46 V
n = 4
Q = [Fe3+]^4/ [Fe2+]^4
Q = [1.9]^4/[2.0]^4
Q = 0.8
Substituting values;
Ecell = 0.46 - 0.0592/4 log (0.8)
Ecell = 0.451 V
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