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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
MCQ
Identify the reaction from following having top position in $EMF$ series (Std. red. potential) according to their electrode potential at $298 \;K$.
  • A
    $K ^{+}+1 e ^{-} \rightarrow K _{( s )}$
  • B
    $Mg ^{2+}+2 e ^{-} \rightarrow Mg _{( s )}$
  • C
    $Fe ^{2+}+2 e ^{-} \rightarrow Fe _{( s )}$
  • ✓
    $Au ^{3+}+3 e ^{-} \rightarrow Au _{( s )}$
✓

Answer

Correct option: D.
$Au ^{3+}+3 e ^{-} \rightarrow Au _{( s )}$
d
$\begin{array}{ll}A u^{3+}+3 e^{-} \rightarrow A u(s) & E^{0}=1.40 \vee \\ F e^{2+}+2 e^{-} \rightarrow F e(s) & E^{0}=-0.44 v \\ M g^{2+}+2 e^{-} \rightarrow M g(s) & E^{0}=-2.36 v \\ K^{+}+1 e^{-} \rightarrow K(s) & E^{0}=-2.93 V\end{array}$

As per electrochemical series, $Au ^{3+}$ occupies. the top position.

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