For a spontaneous reaction the ΔG, equilibrium constant (K) and E°cell will be respectively :

Question

For a spontaneous reaction the ΔG, equilibrium constant (K) and E°cell will be respectively : (A) -ve, > 1, -ve (B) -ve, < 1, -ve (C) +ve, > 1, -ve (D) -ve, > 1, +ve

Tardigrade Admin · Accepted Answer

Δ G° = -2.303RT log Keq Δ G° = -nFE°cell If a cell reaction is spontaneous (proceeding in forward side), it means Keq > 1 and E° cell = +ve Thus Δ G° = - ve

Q. For a spontaneous reaction the $Δ$ G, equilibrium constant (K) and $E_{ce ll}^{\circ}$ will be respectively :

$- v e, > 1, - v e$

$- v e, < 1, - v e$

$+ v e, > 1, - v e$

$- v e, > 1, + v e$

$Δ G^{\circ} = - 2.303 RT l o g K_{e q}$
$Δ G^{\circ} = - n F E_{ce ll}^{\circ}$
If a cell reaction is spontaneous (proceeding in forward side), it means
$K_{e q} > 1$ and $E_{ce ll}^{\circ} = + v e$
Thus $Δ G^{\circ} = - v e$

Q. For a spontaneous reaction the ΔG, equilibrium constant (K) and Ecell∘​ will be respectively :

−ve,>1,−ve

−ve,<1,−ve

+ve,>1,−ve

−ve,>1,+ve