The standard Gibbs energy change -triangle G-circ- in kJ mol-1- in a Daniel cell -E-cell-circ- - 1-1V- when 2 moles of Zn-s- is oxidized 298 K- is closest to-212-3-106-2-424-6-53-1

Question

Toppr Admin · Accepted Answer

-x394-G-x2212-nFEocell-xA0- -xA0- -xA0- -xA0- -x2212-2-xD7-96500-xD7-1-1-x2212-212-3-xA0-KJHence- option C is correct-

The standard Gibbs free energy change $(△ G^{\circ}$ in $k J m o l^{- 1})$ , in a Daniel cell $(E_{c e l l}^{\circ} = 1.1 V)$ , when $2$ moles of $Z n (s)$ is oxidized at $298 K$ , is closest to:
$- 212.3$
$- 106.2$
$- 424.6$
$- 53.1$

$Δ G = - n F E_{c e l l}^{o}$

$= - 2 \times 96500 \times 1.1 = - 212.3 K J$

Hence, option C is correct.

The standard Gibbs free energy change (△G∘ in kJ mol−1), in a Daniel cell (E∘cell=1.1V), when 2 moles of Zn(s) is oxidized at 298 K, is closest to:−212.3−106.2−424.6−53.1

ΔG=−nFEocell =−2×96500×1.1=−212.3 KJHence, option C is correct.

The standard Gibbs free energy change $(△ G^{\circ}$ in $k J m o l^{- 1})$ , in a Daniel cell $(E_{c e l l}^{\circ} = 1.1 V)$ , when $2$ moles of $Z n (s)$ is oxidized at $298 K$ , is closest to:
$- 212.3$
$- 106.2$
$- 424.6$
$- 53.1$

$Δ G = - n F E_{c e l l}^{o}$

$= - 2 \times 96500 \times 1.1 = - 212.3 K J$

Hence, option C is correct.