Question

JEE MAINS CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

Gibbs energy ( ∆G ) is negative if

$\sf \to \: c_{2} = \dfrac{c_{1} }{ \sqrt{2} } = option \: (1) \: is \: correct$

EXPLANATION.

$\sf \to \: cu(s) |cu {}^{2 + }( c_{2}m) || {cu}^{2 + } ( c_{1}m) |cu(s) \\ \\ \sf \to \: for \: concentration \: cell \: e {}^{0} _{cell} = 0$

$\sf \to \: anode \implies \: cu(s) \longrightarrow \: cu {}^{2 + } + (aq)a \\ \\ \sf \to \: cathode \implies \: cu {}^{2 + } + (aq)c \longrightarrow \: cu(s) \\ \\ \sf \to \: overall \: net \: reaction \: or \: \: redox \: \: reaction \\ \\ \sf \to \: cu {}^{2 + } + (aq)c \longrightarrow \: cu {}^{2 + } + (aq)a$

$\sf \to \: as \: we \: know \: that \: \Delta \: g \: = -nf e {}^{0} _{cell} \\ \\ \sf \to \: \Delta \: g \: = - ve \: then \: e {}^{0} _{cell} \: is \: positive \\ \\ \sf \to \: as \: we \: know \: that \: the \: nerst \: equation \\ \\ \sf \to \: e_{cell} = e {}^{0} _{cell} - \frac{0.529}{2} log( \frac{r}{p} )$

$\sf \to \: e_{cell} = - \frac{0.529}{2} log( \dfrac{ c_{2} }{ c_{1} } ) \\ \\ \sf \to \: e_{cell} > 0 \\ \\ \sf \to \: c_{2} < c_{1} \\ \\ \sf \to \: c_{2} = \frac{ c_{1} }{ \sqrt{2} } = answer$