The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is: Zn(s) + Cu2+ (aq) arrow Zn2+ (aq) + Cu (s), E° = 2 V at 298 K (Faraday's constant, F = 96000 C mol-1)

Question

The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is: Zn(s) + Cu2+ (aq) arrow Zn2+ (aq) + Cu (s), E° = 2 V at 298 K (Faraday's constant, F = 96000 C mol-1) (A) -384 (B) -192 (C) 192 (D) 384

Tardigrade Admin · Accepted Answer

Δ G° = -nFE°cell =-2 × 96000 × 2 = -384000 J = -384 kJ

Q. The standard Gibbs energy for the given cell reaction in $k J m o l^{- 1}$ at $298 K$ is : $Z n (s) + C u^{2} + (a q) \to Z n^{2 +} (a q) + C u (s), E ° = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C m o l^{- 1}$ )

-384

-192

192

384

$Δ G^{\circ} = - n F E_{ce ll}^{\circ}$
= $- 2 \times 96000 \times 2$
= $- 384000 J$
= $- 384 k J$

Q. The standard Gibbs energy for the given cell reaction in kJmol−1 at 298K is : Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s),E°=2V at 298K (Faraday's constant, F=96000Cmol−1)

-384

-192

192

384

ΔG∘=−nFEcell∘​ =−2×96000×2 = −384000J = −384kJ

Q. The standard Gibbs energy for the given cell reaction in $k J m o l^{- 1}$ at $298 K$ is : $Z n (s) + C u^{2} + (a q) \to Z n^{2 +} (a q) + C u (s), E ° = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C m o l^{- 1}$ )

$Δ G^{\circ} = - n F E_{ce ll}^{\circ}$
= $- 2 \times 96000 \times 2$
= $- 384000 J$
= $- 384 k J$