Answer:
The [tex]E_{cell}^{o}[/tex] for the reaction is 1.14V.
Explanation:
The given chemical reaction is as follows.
[tex]CH_{3}CH_{2}OH(l)+3O_{2}(g) \rightarrow 2CO_{2}(g)+3H_{2}O(l)[/tex]
The [tex]\Deta G[/tex] can be calculated by the following formula.
[tex]\Delta G= \Delta G(products)-\Delta G(reactants)[/tex]
[tex]=(3\times -237.129kJ)+(2\times -394.359kJ)-(-174.8kJ)-0J[/tex]
[tex]=-1325.30kJ[/tex]
[tex]\Delta G= -nFE_{cell}^{o}[/tex]
Rearrange the formula is as follows
[tex]E_{cell}^{o}=\frac{\Delta G}{-nF}[/tex]
n= Number of moles of electrons transferred.
[tex]=2\times 16= 12\,electrons[/tex]
[tex]=\frac{-1325.305\times 1000J}{-12\times96485}=1.14V[/tex]
Therefore, The [tex]E_{cell}^{o}[/tex] for the reaction is 1.14V.
The value of E ∘ cell for this cell at 25 °C is 1.14V.
Given the chemical reaction:
CH 3 CH 2 OH ( l ) + 3 O 2 ( g ) ⟶ 2 CO 2 ( g ) + 3 H 2 O ( l )
To find the G,
△G= △G(products) - △G(reactants)
= (3 x -237.129KJ) + (2 x -394.359KJ) - (-174.8KJ) - 0J
=-1325.30KJ.
Hence, to find the value of E ∘ cell,
= -1325.305 x 1000J/-12 x 96485
= 1.14V
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