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Chapter 1: Problem 22

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

Short Answer

Expert verified
The reaction potential for the given reaction is -0.7 V. Hence, the correct answer is (A) -0.7 V.

Step by step solution

01

Identify the Oxidation and Reduction Components

In the reaction, \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \(\mathrm{Fe}+\mathrm{Cu}^{2+}\), Iron (Fe) is being reduced (gaining electrons) while Copper (Cu) is being oxidised (losing electrons).
02

Write Down the Reduction Potentials

The reduction potential for Fe is -0.4 V and for Cu is +0.3 V. However, since Cu is getting oxidized, its sign needs to be reversed, making it -0.3 V.
03

Apply the Formula

Now, use the formula: E(reaction) = E(oxidation) + E(reduction). Here, E(oxidation) is for Cu which is -0.3 V, and E(reduction) is for Fe which is -0.4 V. Substituting these into the formula gives: E(reaction) = -0.3 + (-0.4) = -0.7 V.

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Most popular questions from this chapter

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)
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