The complete question is ;
Calculate the equilibrium constant, K, for the reaction shown at 25 °C.
Fe3+(aq)+B(s)+6H2O(l)-------->Fe(s)+H3BO3(s)+3H3O+(aq)
The balanced reduction half?reactions for the equation and their respective standard reduction potential values (E?) are
Fe3+(aq)+3e------>Fe(s) E°=-0.04V
H3BO3(s)+3H3O+(aq)+3e------->B(s)+6H2O(l) E°=0.8698 V
Answer:
1.05
Explanation:
E° cell= E°cathode- E°anode
E°cathode = 0.8698 V
E°anode= -0.04V
E°cell=0.8698 V - (-0.04V)
E°cell= 0.9098 V
E°cell= 0.0592/n logK
From the balanced redox reaction equations above, n=3
0.9098 V= 0.0592/3 logK
logK= 0.0217
K= Antilog (0.0217)
K= 1.05
The equilibrium constant for the given reaction is 1.05.
Given here,
The reaction
[tex]\bold {Fe_3+(aq)+B(s)+6H_2O(l)\rightarrow Fe(s)+H_3BO_3(s)+3H_3O+(aq)}[/tex]
The formula of equilibrium constant,
[tex]\bold{E^ocell= E^ocathode- E^oanode}[/tex]
Where,
[tex]\bold{ E^ocathode}[/tex] = 0.8698 V
[tex]\bold{E^oanode}[/tex] = -0.04V
Put the values in the formula,
[tex]\bold{E^ot= 0.8698 V- -0.04V}\\\\\bold{E^ot= 0.9098 V }[/tex]
[tex]\bold {E^ot= \dfrac {0.0592}{n logK}}[/tex]
Solve the equation for K, we get
[tex]\bold{k = 1.05 }[/tex]
Therefore, the equilibrium constant for the given reaction is 1.05.
To know more about equilibrium constant, refer to the link:
https://brainly.com/question/10038290
