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Chapter 1: Problem 20

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_{2}+7 \mathrm{H}_{2} \mathrm{O}\) Which of the following statements about the reaction given above is NOT true? (A) The oxidation number of chromium changes from \(+6\) to \(+3 .\) (B) The oxidation number of iodine changes from \(-1\) to 0. (C) The oxidation number of hydrogen changes from +1 to 0. (D) The oxidation number of oxygen remains the same.

Short Answer

Expert verified
The statement in option (C) is not true because the oxidation number of Hydrogen does not change from +1 to 0 during the reaction.

Step by step solution

01

Oxidation Number of Chromium

Looking into option (A), the oxidation of Chromium (Cr) in\(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) is +6. This is because Oxygen always has an oxidation number of -2 in its compounds, so the two Chromium must have a total oxidation number of +12 for the total charge to be -2. In the products, each \(\mathrm{Cr}^{3+}\) atom has an oxidation state of +3. So, Chromium's oxidation state changes from +6 in the reactants to +3 in the products. Option (A) is therefore true.
02

Oxidation Number of Iodine

Option (B): Iodine starts with an oxidation number of -1 in \(\mathrm{I}^{-}\) and changes to 0 in \(\mathrm{I}_{2}\). Therefore, the statement in option (B) is correct.
03

Oxidation Number of Hydrogen

Option (C) suggests that Hydrogen changes oxidation state from +1 to 0, which is incorrect. In both the reactants and products, Hydrogen is part of a compound (first being \(\mathrm{H}^{+}\) and later being in \(\mathrm{H}_{2}\mathrm{O}\)) and has an oxidation state of +1. Hydrogen's oxidation state does not change in this reaction, which makes option (C) false.
04

Oxidation Number of Oxygen

Finally checking, option (D), the oxidation state of Oxygen does not change in this reaction. It stays at -2, as it is in both the reactants (\(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\)) and the products (\(\mathrm{H}_{2} \mathrm{O} \)). Therefore, this statement is also true.

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Most popular questions from this chapter

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. Identify the correct net ionic equation for the reaction that takes place. (A) \(\mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q) \mapsto \mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (B) \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \leftrightarrow \mathrm{NaCN}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (D) \(\mathrm{H}^{+}(a q)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}\) (aq)

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ What will happen in the salt bridge as the reaction progresses? (A) The Na' ions will flow to the Cu/Cu' half-cell. (B) The Br' ions will flow to the Cu/Cu' half-cell. (C) Electrons will transfer from the Cu/Cu' half-cell to the Zn/Zn" half cell. (D) Electrons will transfer from the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell to the Cu/Cu' half cell.
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