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