Answer:
Δ[tex]G^{o}[/tex] = -461198.3 J
Explanation:
Gibbs free energy is defined as the energy associated with a given chemical reaction which can be used to do work. Firstly, we need to figure out the chemical equation for the given problem. For the given problem, the chemical equation is:
[tex]Mn^{2+} _{(aq)} + 2e^{-}[/tex]⇒[tex]Mn_{(s)}[/tex] [tex]E^{o} = -1.18 V[/tex]
[tex]Mn^{2+} _{(aq)} + H_{2}O[/tex] ⇒[tex]MnO_{2(s)} +4H^{+} + 2e^{-}[/tex][tex]E^{o} = -1.21 V[/tex]
The addition of the two [tex]E^{o}[/tex] gives the [tex]E_{cell}[/tex] of the equation, i.e. -1.18-1.21 = -2.39 V.
Then, using the equation for Δ[tex]G^{o}[/tex], we have:
Δ[tex]G^{o}[/tex] = n*F*[tex]E^{o}[/tex] = 2*96485*-2.39 = -461198.3 J
Based on the calculations, the Gibbs's free energy is equal to -461,198.3 Joules.
Scientific data:
Gibbs's free energy can be defined as the quantity of energy that is associated with a chemical reaction.
First of all, we would write a properly balanced chemical equation for this chemical reaction:
[tex]Mn^{2+}(aq)+2e^{-} \rightarrow Mn (s) \;;E^{0}=-1.18V\\\\Mn^{2+}(aq)+H_2O \rightleftharpoons MnO_2(s) +H^{+}+2e^{-} \;;E^{0}=-1.21V[/tex]
For the electromotive force:
[tex]E^{0}_{cell}=-1.18-1.21 = -2.39[/tex]
Mathematically, Gibbs's free energy is given by the formula:
[tex]\Delta G ^{0}=-nFE^{0}_{cell}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]\Delta G ^{0}=-2 \times 96485(-2.39)\\\\\Delta G ^{0}=-461198.3 \;Joules.[/tex]
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