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Chapter 1: Problem 5

A strip of metal \(\mathrm{X}\) is placed into a solution containing \(\mathrm{Y}^{2+}\) ions and no reaction occurs. When metal \(\mathrm{X}\) is placed in a separate solution containing \(\mathrm{Z}^{2+}\) ions, metal \(\mathrm{Z}\) starts to form on the strip. Which of the following choices organizes the reduction potentials for metals \(\mathrm{X}, \mathrm{Y},\) and \(\mathrm{Z}\) from greatest to least? (A) \(\quad \mathrm{X}>\mathrm{Y}>\mathrm{Z}\) (B) \(\quad \mathrm{Y}>\mathrm{Z}>\mathrm{X}\) (C) \(\quad \mathrm{Z}>\mathrm{X}>\mathrm{Y}\) (D) \(\mathrm{Y}>\mathrm{X}>\mathrm{Z}\)

Short Answer

Expert verified
The reduction potentials for metals X, Y and Z are organized from greatest to least as Y>X>Z, i.e., Choice (D).

Step by step solution

01

Analyze the reactivity of metal X with Y ions

The provided information states that no reactions occur when a strip of metal X is put into a solution containing Y ions. This implies that metal X cannot be oxidized by Y ions or in other words, Y2+ ions have lower reduction potential than metal X.
02

Analyze the reactivity of metal X with Z ions

When metal X is dipped in a solution of Z ions, it is recorded that metal Z starts to form. This implies that metal X is oxidizing Z2+ ions. That means Z2+ ions have higher reduction potential than metal X.
03

Arrange the reduction potentials

Based on these observations, it is concluded that Y has the highest reduction potential followed by X and lastly Z, because Y ions couldn't be reduced by X and Z ions were reduced by X. Therefore, the order of reduction potential from greatest to least will be \(Y>X>Z\). Comparing this conclusion with the four options given in the exercise, the correct choice is (D).

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Most popular questions from this chapter

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