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The spontaneous redox reaction/{s} among the follow is/ are {(

Question

The spontaneous redox reaction $/ s$ among the follow is $/$ are
$(a)$ $2 F e^{3 +} + F e \to 3 F e^{++}$
$(b) H g_{2}^{++} \to H g^{++} + H g$
$(c)$ $3 A g C l + N O + 2 H_{2} O \to 3 A g + 3 C l^{-} + N O_{3}^{-} + 4 H^{+}$
Given that
\displaystyle{{E}_{{{F}{e}{\left(+++\right)}/{F}{e}^{{++}}}}^{{.}}}={0.77}{V}\ \text{ }\ {{E}_{{{F}{e}^{{++}}/{F}{e}}}^{{.}}}=-{0.44}{V}
\displaystyle{{E}_{{{H}{g}^{{\pm}}/{H}{g}}}^{{.}}}={0.85}{V}\ \text{ }\ {{E}_{{{H}{g}^{{++}}/{H}{g}_{{{2}}}}}^{{.}}}={0.92}{V}
\displaystyle{{E}_{{{A}{g}{C}{l}/{A}{g}}}^{{.}}}={0.22}{V}\ \text{ }\ {{E}_{{{N}{O}_{{{3}}}/{N}{O}}}^{{.}}}={0.96}{V}

a

a, b, c

a, b

a, c

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Sebastian Discussed

The spontaneous redox reaction

/ s

among the follow is

/

are

(a)

2 F e^{3 +} + F e \to 3 F e^{++}

(b) H g_{2}^{++} \to H g^{++} + H g

(c)

3 A g C l + N O + 2 H_{2} O \to 3 A g + 3 C l^{-} + N O_{3}^{-} + 4 H^{+}

Given that
\displaystyle{{E}_{{{F}{e}{\left(+++\right)}/{F}{e}^{{++}}}}^{{.}}}={0.77}{V}\ \text{ }\ {{E}_{{{F}{e}^{{++}}/{F}{e}}}^{{.}}}=-{0.44}{V}
\displaystyle{{E}_{{{H}{g}^{{\pm}}/{H}{g}}}^{{.}}}={0.85}{V}\ \text{ }\ {{E}_{{{H}{g}^{{++}}/{H}{g}_{{{2}}}}}^{{.}}}={0.92}{V}
\displaystyle{{E}_{{{A}{g}{C}{l}/{A}{g}}}^{{.}}}={0.22}{V}\ \text{ }\ {{E}_{{{N}{O}_{{{3}}}/{N}{O}}}^{{.}}}={0.96}{V}

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Question Text	The spontaneous redox reaction $/ s$ among the follow is $/$ are $(a)$ $2 F e^{3 +} + F e \to 3 F e^{++}$ $(b) H g_{2}^{++} \to H g^{++} + H g$ $(c)$ $3 A g C l + N O + 2 H_{2} O \to 3 A g + 3 C l^{-} + N O_{3}^{-} + 4 H^{+}$ Given that \displaystyle{{E}_{{{F}{e}{\left(+++\right)}/{F}{e}^{{++}}}}^{{.}}}={0.77}{V}\ \text{ }\ {{E}_{{{F}{e}^{{++}}/{F}{e}}}^{{.}}}=-{0.44}{V} \displaystyle{{E}_{{{H}{g}^{{\pm}}/{H}{g}}}^{{.}}}={0.85}{V}\ \text{ }\ {{E}_{{{H}{g}^{{++}}/{H}{g}_{{{2}}}}}^{{.}}}={0.92}{V} \displaystyle{{E}_{{{A}{g}{C}{l}/{A}{g}}}^{{.}}}={0.22}{V}\ \text{ }\ {{E}_{{{N}{O}_{{{3}}}/{N}{O}}}^{{.}}}={0.96}{V}
Answer Type	Video solution: 1
Upvotes	150

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