The relation between standard free energy - G- and standard emf E- cell is Given - n is the number of electrons involved and F is Faraday constant

The relation between standard free energy (Δ G^∘) and standard emf (E^∘ _cell) is nbsp; nbsp;Δ G^∘= nFE^∘ _cell ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibb's Energy and Nernst Equation

The relation ...

Question

The relation between standard free energy $(Δ G^{\circ})$ and standard emf $(E_{c e l l}^{\circ})$ is
(Given : $n$ is the number of electrons involved and $F$ is Faraday constant)

Δ G^{\circ} = n F E_{c e l l}^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ G^{\circ} = - n F E_{c e l l}^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ G^{\circ} = \frac{n F}{E_{c e l l}^{\circ}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ G^{\circ} = - \frac{n F}{E_{c e l l}^{\circ}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

F

e

N

Δ G = Change in Gibbs' free energy n = no. of moles of electron transferred in a reaction F = Faraday constant E_{c e l l} = Cell potential or emf

(K)

(Δ G^{o})

(T)

(- Δ G)

- Δ G = - Δ H - T {[\frac{δ (Δ G)}{δ T}]}_{P}

- Δ H

E = \frac{- Δ H}{n F} + T {[\frac{δ E}{δ T}]}_{P}

Δ H, Δ G

Δ S

E^{0} = \frac{2.303 R T l o g k}{n F}

\begin{matrix} Z n & Z n^{+ 2} & F e^{+ 3} F e^{+ 2} (S) & (a q) & (a q) (a q) \end{matrix}

25^{\circ}

50^{\circ}

Δ S^{\circ}

Join BYJU'S Learning Program