Electrochemistry - Result Question 56
####5. Consider the following reduction processes:
$$ \begin{aligned} & Zn^{2+}+2 e^{-} \longrightarrow Zn(s) ; E^{\circ}=-0.76 V \ & Ca^{2+}+2 e^{-} \longrightarrow Ca(s) ; E^{\circ}=-2.87 V \ & Mg^{2+}+2 e^{-} \longrightarrow Mg(s) ; E^{\circ}=-2.36 V \ & Ni^{2+}+2 e^{-} \longrightarrow Ni(s) ; E^{\circ}=-0.25 V \end{aligned} $$
The reducing power of the metals increases in the order
(2019 Main, 10 Jan I)
(a) $Zn<Mg<Ni<Ca$
(b) $Ni<Zn<Mg<Ca$
(c) $Ca<Zn<Mg<Ni$
(d) $Ca<Mg<Zn<Ni$
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Answer:
Correct Answer: 5. (b)
Solution:
- Reducing power of an element
$$ \propto \frac{1}{\text { Standard reduction potential }} $$
Here, $E^{\circ}{ } _{M^{2+} / M}$ values of the given metals are as,
$$ \begin{array}{ccccc} \text { Metals } & Ni & Zn & Mg & Ca \ \boldsymbol{E}^{\circ}(V) & -0.25 & -0.76 & -2.36 & -2.87 \ & & & \ \hline \end{array} $$
Thus, the correct order of increasing reducing power of the given metal is,
$$ Ni<Zn<Mg<Ca $$
