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Chapter 17: Problem 103

It took 2.30 min using a current of 2.00 A to plate out all the silver from 0.250 L of a solution containing Ag \(^{+} .\) What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

Short Answer

Expert verified
The original concentration of Ag+ ions in the solution was approximately 0.01144 mol/L.

Step by step solution

01

Calculate the Charge (Coulombs) Passed Through the Cell

We are given that a current of 2.00 A flowed for 2.30 minutes. The total charge in Coulombs can be calculated using the formula: Charge (Q) = Current (I) × Time (t) Now we need to convert the time from minutes to seconds, as the unit of current (Ampere) is in terms of seconds. 1 minute = 60 seconds, so 2.30 minutes = 2.30 × 60 seconds = 138 seconds Now, we can find the total charge: Q = I × t Q = 2.00 A × 138 seconds Q = 276 Coulombs
02

Calculate the Moles of Ag+ ions Deposited

Next, we will use Faraday's laws of electrolysis to find the moles of Ag+ ions deposited. The relationship between the charge, the number of moles of substance, and the Faraday's constant is: Charge (Q) = Number of moles (n) × Faraday's constant (F) F is the Faraday constant, equal to 96485 C/mol In this case, the deposited substance is silver (Ag), which is a monovalent ion (Ag+), so the Faraday's law for Ag+ can be represented as: Q = n × F Now, rearrange the formula to find the number of moles: Number of moles (n) = Charge (Q) / Faraday's constant (F) n = 276 C / 96485 C/mol n ≈ 0.00286 mol
03

Calculate the Original Concentration of Ag+ ions

Finally, to find the original concentration of Ag+ ions in the solution, we will use the formula: Concentration (c) = Number of moles (n) / Volume (V) We are given that the volume of the solution is 0.250 L. Using the moles of Ag+ ions calculated in Step 2, we can find the concentration: c = 0.00286 mol / 0.250 L c ≈ 0.01144 mol/L So, the original concentration of Ag+ ions in the solution was approximately 0.01144 mol/L.

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Most popular questions from this chapter

The compound with the formula TII \(_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{T}^{3+}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Tl}^{+} & & \mathscr{E}^{\circ}=1.25 \mathrm{V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} & \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{E}^{\circ}=0.55 \mathrm{V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12} .\) Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$

Specify which of the following equations represent oxidationreduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.
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