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Chemistry
Posted 4 months ago

An experiment was performed to investigate the reaction between Ln\mathrm{Ln} metal and Ni2+(aq)\mathrm{Ni}^{2+}(a q) at different concentrations. Because the Ni2+(aq)\mathrm{Ni}^{2+}(a q) ion is green, the extent of the reaction was determined using spectrophotometric analysis. Four 20.0 mL20.0 \mathrm{~mL} standard solutions of Ni2+(aq)\mathrm{Ni}^{2+}(a q) were prepared by dissolving NiCl26H2O\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} (molar mass 240 g/mol240 \mathrm{~g} / \mathrm{mol} ) in water. The absorbance of each solution was measured; the results are shown both in the table below and in the following plot of the absorbance data.
\begin{tabular}{|c|c|c|}
\hline Solution & {[Ni2+]\left[\mathrm{Ni}^{2+}\right]} & Absorbance \\
\hline 1 & 0.020 & 0.12 \\
\hline 2 & 0.040 & 0.25 \\
\hline 3 & 0.060 & 0.39 \\
\hline 4 & 0.080 & 0.51 \\
\hline
\end{tabular}
15. Which species is being oxidized in the reaction between Zn\mathrm{Zn} metal and NiCl2(aq)\mathrm{NiCl}_{2}(a q) ?
(A) Zn\mathrm{Zn}
(B) Zn2+\mathrm{Zn}^{2+}
(C) Ni\mathrm{Ni}
(D) Ni2+\mathrm{Ni}^{2+}
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Answer from Sia
Posted 4 months ago
Solution
1
Identify the species involved in oxidation and reduction: In a redox reaction, the species that gets oxidized loses electrons, while the species that gets reduced gains electrons
2
Analyze the possible reactions: Zn\mathrm{Zn} can be oxidized to Zn2+\mathrm{Zn}^{2+} by losing two electrons (ZnZn2++2e\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^{-}). Ni2+\mathrm{Ni}^{2+} can be reduced to Ni\mathrm{Ni} by gaining two electrons (Ni2++2eNi\mathrm{Ni}^{2+} + 2e^{-} \rightarrow \mathrm{Ni})
3
Determine the species being oxidized: Since Zn\mathrm{Zn} metal can lose electrons to form Zn2+\mathrm{Zn}^{2+}, it is the species that is being oxidized in the reaction with NiCl2(aq)\mathrm{NiCl}_{2}(aq)
Answer
(A) Zn\mathrm{Zn}
Key Concept
Oxidation in a redox reaction
Explanation
In the reaction between Zn\mathrm{Zn} metal and NiCl2(aq)\mathrm{NiCl}_{2}(aq), Zn\mathrm{Zn} is oxidized to Zn2+\mathrm{Zn}^{2+} as it loses electrons.

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