Answer :
Answer:
Emf of the cell is 0.335 V
Explanation:
[tex]4Fe^{2+}(aq)+O_{2}(g)+4H^{+}(aq)\rightarrow 4Fe^{3+}+2H_{2}O(l)[/tex]
[tex]E = E^{0}-\frac{RT}{nF}ln\frac{[Product]}{[Reactant]}[/tex]
[tex]E = E^{0}-2.303\frac{RT}{nF}log\frac{[Product]}{[Reactant]}[/tex]
[tex]E = E^{0}-\frac{0.059}{n}log\frac{[Product]}{[Reactant]}[/tex]
Here ,
n = number of electrons transferred
[tex]E^{0} = E_{cathode}- E_{anode}[/tex]
Cathode : Reduction-Reaction
[tex]O_{2}(g)+4H^{+} +4e^{-}(aq)\rightarrow 2H_{2}O(l)[/tex].......1.229 V
Anode :Oxidation Reaction
[tex]4Fe^{2+}\rightarrow Fe^{3+} + e^{-}[/tex].........0.771 V
The electron are balanced by multiplying Anode with 4(n= 4)
In this Reaction , Fe is oxidized from +2 to +3 state and oxygen is reduced to H2O
[tex]E^{0} = E_{O_{2}/Water}- E_{Fe^{3+}/Fe^{2+}}[/tex]
[tex]E_{O_{2}/Water}[/tex] = +1.229 V (from standard reduction potential table)
[tex]E_{Fe^{3+}/Fe^{2+}}[/tex]= 0.771 V
[tex]E^{0}[/tex] = 1.229 - 0.771 = 0.458 V
Insert the value in Nernst Equation
E = 0.458 V
n =4
[tex][Fe^{3+}]= 1.1\times 10^{-2}M[/tex] = 0.011 M
[tex][Fe^{2+}]= 1.8M[/tex]
[tex]pO_{2} = 0.43atm[/tex]
pH = 4.2
pH = -log[H+]
[H+] = antilog(-4.2)
[tex][H^{+}]= 6.309\times 10^{-5}M[/tex]
[tex]E = 0.458-\frac{0.059}{4}logK[/tex]
[tex]K =\frac{[Fe^{3+}]^{4}}{[Fe^{2+}]^{4}[H+]^{4}pO_{2}}[/tex]
[tex]E = 0.458-\frac{0.059}{4}logK[/tex]
[tex]K=\frac{(.011^{4}}{(1.8)^{4}(6.309)10^{-5})^{4}0.43}[/tex]
[tex]E=0.458 -\frac{0.059}{4}log\frac{0.011^{4}}{(7.5)10^{-17}}[/tex]
[tex]= 0.458 -\frac{0.059}{4}log(2.047\times 10^{8})[/tex]
[tex]= 0.458 -\frac{0.059}{4}(2.077)[/tex]
[tex]= 0.458 -0.122[/tex]
[tex]=0.335 V[/tex]
Emf of the cell is 0.335 V
Let's write the reaction involved:
[tex]4 Fe^{2+} (aq) + O_2(g)+ 4H^{+} (aq)[/tex] → [tex]4 Fe^{3+} (aq) +2 H_2 O (l)[/tex]
Nernst equation:
[tex]E= E^{0} -\frac{nF}{RT} ln \frac{[Products]}{[Reactants]} \\\\\\E= E^{0} -2.303\frac{nF}{RT} log \frac{[Products]}{[Reactants]}\\\\\\E= E^{0} -\frac{0.059}{n} log \frac{[Products]}{[Reactants]}[/tex]
where n= number of electrons,
[tex]E^{0} = E_{cathode}-E_{anode}[/tex]
Since, oxidation takes place at cathode. Thus the reaction is:
[tex]O_2(g)+ 4H^{+}(aq) + 4 e^-[/tex] → [tex]2 H_2 O (l)[/tex], the value of reduction potential =1.229V
As, reduction takes place at anode. Thus the reaction will be:
[tex]4 Fe^{3+}[/tex] → [tex]Fe^{3+} (aq) + 4e^-[/tex], the value of reduction potential =0.771 V
In this Reaction , Fe is oxidized from +2 to +3 state and oxygen is reduced to H₂O.
[tex]E^0 =E_{O_2/water}-E_{Fe^{3+}/Fe^{2+}}\\\\E_{O_2/water}=+1.2299 V\\\\E_{Fe^{3+}/Fe^{2+}}=0.771V\\[/tex]
[tex]E^0\\[/tex]= 1.229 - 0.771 = 0.458 V
Insert the value in Nernst Equation
E = 0.458 V
n =4
It is given that:
[tex][Fe^{3+}]=0.011 M\\ [Fe^{2+}]=1.8 M[/tex]
pO₂=0.43 atm
pH = 4.2
pH = -log[tex][H^+][/tex]
[tex][H^+][/tex] = antilog(-4.2)
[tex][H^+]=6.309*10^{-5} M[/tex]
[tex]E=0.458-\frac{0.059}{4} log K\\[/tex]
The value of K can be calculated as [tex]K=\frac{[Fe^{3+}]^4}{[Fe^{2+}]^4 [H^+]^4 pO_2}[/tex]
On substituting the values,
[tex]E=0.458 -\frac{0.059}{4} log \frac{0.011^4 }{7.5*10^{-17}} \\\\=0.458 -\frac{0.059}{4} log (2.047 * 10^8)\\\\=0.458-0.122\\=0.335 V[/tex]
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