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Gibb's Energy and Nernst Equation

Which one is ...

Question

Which one is correct relation?

Δ G = Δ G^{o} + R T l n Q

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- Δ G^{o} = R T l n K

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Δ H = n F [T {(\frac{\partial E}{\partial T})}_{P} - E]

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E^{o} = \frac{R T}{n F} l n K

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Solution

The correct options are
A $Δ G = Δ G^{o} + R T l n Q$
B $- Δ G^{o} = R T l n K$
C $Δ H = n F [T {(\frac{\partial E}{\partial T})}_{P} - E]$
D $E^{o} = \frac{R T}{n F} l n K$
A. The relationship between the free energy change and the standard free energy change is given by the expression $Δ G = Δ G^{o} + R T l n Q$ .
Thus, the option A is correct.
B. At equilibrium, the value of the free energy change is zero. Hence, the expression becomes $- Δ G^{o} = R T l n K$ .
Thus, the option B is correct.
C. The relationship between the enthalpy change and the cell potential is $Δ H = n F [T {(\frac{\partial E}{\partial T})}_{P} - E]$ .
Thus, the option C is correct.
D. The relationship between the cell potential and the equilibrium constant is $E^{o} = \frac{R T}{n F} l n K$ .
Thus, the option D is correct.

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Similar questions

E^{o}

Δ G^{o}

E^{o} = - Δ G^{o} / n F

A g^{+}

A g^{+} + e^{-} \to A g

Δ G^{o} = - n F E_{c e l l}^{o}

Δ G^{o} = - R T ln k

n = 2

(K)

(Δ G^{o})

(T)

E = E^{o} - \frac{R T}{n F} l n Q

Q = K_{c}

⇌

Δ G^{o} = 250 k J m o l^{- 1}

⇌

Δ G^{o} = - 100 k J m o l^{- 1}

⇌

Δ G^{o} = - 150 k J m o l^{- 1}

⇌

Δ G^{o} = 150 k J m o l^{- 1}

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