Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.
1 answer:
Following reaction occurs in the given electrochemical system:
→ Fe + 
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where,
= standard reduction potential of Fe2+/Fe = -0.44 v
= standard reduction potential of Zn2+/Zn = -0.763 v
E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, Δ
On equating and rearranging equation 1 and 2, we get
K = exp(
)= exp (
) = 8.46 x 
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where,
E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, Δ
On equating and rearranging equation 1 and 2, we get
K = exp(
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Answer : The value of is,
Explanation :
The formula used for is:
............(1)
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R = gas constant = 8.314 J/mole.K
T = temperature =
Q = reaction quotient
First we have to calculate the .
Formula used :
Now put all the given values in this formula, we get:
Now we have to calculate the value of 'Q'.
The given balanced chemical reaction is,
The expression for reaction quotient will be :
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
Now we have to calculate the value of by using relation (1).
Now put all the given values in this formula, we get:
Therefore, the value of is,
Answer:
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Explanation:
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where,
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b is Wien's displacement constant (2.89×10⁻³ m⋅K)
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