JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

The standard Gibbs energy for the given cell reaction in kJ mol^–1 at 298 K is

Zn(s) + Cu²⁺(aq) $\to$ Zn²⁺(aq) + Cu(s),

E° = 2 V at 298 K

(Faraday’s constant, F = 96000 C mol^–1)

Solution

$Δ G^{0} = - nFE_{Cell}^{o}$

$= - 2 \times (96000) \times 2 V$

$= - 384 kJ/mole$ .

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.