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Chapter 20: Problem 72

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) are consumed at the anode of the battery. What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge?

Short Answer

Expert verified
During the discharge of the alkaline battery, 11.96 g of MnO2 is reduced at the cathode when 4.50 g of Zn is consumed at the anode.

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation representing the overall reaction can be written as: \[Zn + 2MnO_{2} + 2H_{2}O \rightarrow Zn(OH)_{2} + 2Mn(OH)_{2}\]
02

Calculate the amount of Zn in moles

We need to convert the mass of Zn (4.50 g) into moles, for which we will use the molar mass of Zn. The molar mass of Zn is 65.38 g/mol, thus: Moles of Zn = \(\frac{4.50\,g}{65.38\,g/mol}\) = 0.0688 mol
03

Determine the stoichiometric ratio between Zn and MnO2

In the balanced chemical equation given above: 1 mol of Zn reacts with 2 mol of MnO2 This gives the stoichiometric ratio between Zn and MnO2 as 1:2.
04

Calculate the amount of MnO2 in moles

Use the stoichiometric ratio to determine the moles of MnO2 reduced in the reaction when 0.0688 mol of Zn is consumed: Moles of MnO2 = \(2 \times 0.0688\,mol\) = 0.1376 mol
05

Calculate the mass of MnO2 reduced

Convert the moles of MnO2 into grams, using the molar mass of MnO2 (86.94 g/mol): Mass of MnO2 = \(0.1376\,mol \times 86.94\,g/mol\) = 11.96 g So during the discharge of the alkaline battery, 11.96 g of MnO2 is reduced at the cathode when 4.50 g of Zn is consumed at the anode.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation is crucial in stoichiometry as it tells us the exact proportion of reactants and products involved in a chemical reaction. For the discharge of an alkaline battery, the balanced equation helps us understand how zinc (Zn) and manganese dioxide (MnO_{2}) interact:\[Zn + 2MnO_{2} + 2H_{2}O \rightarrow Zn(OH)_{2} + 2Mn(OH)_{2}\]This equation shows that one mole of zinc reacts with two moles of manganese dioxide and two moles of water to produce zinc hydroxide and manganese hydroxide. It gives a clear picture of the substance transformation in this electrochemical process.
  • Ensures the same number of each type of atom on both sides.
  • Provides a basis for calculating reactants and products.
Without balancing, any subsequent calculations about quantities in reactions would be inaccurate.
Molar Mass Calculation
Molar mass is a critical concept in converting mass to moles, which is essential in stoichiometric calculations. It is the mass of one mole of a substance, measured in grams per mole (g/mol). For metals like Zn, it is easily found on the periodic table, with zinc having a molar mass of 65.38 g/mol.To find the moles of zinc: \[\text{Moles of Zn} = \frac{4.50 \, \text{g}}{65.38 \, \text{g/mol}}\]evaluates to approximately 0.0688 moles of zinc.
  • The molar mass allows conversion from mass to moles.
  • Essential for determining ratios and quantities in reactions.
Understanding and utilizing the molar mass supports accurate calculations in chemistry.
Stoichiometric Ratio
The stoichiometric ratio extracted from the balanced chemical equation indicates how substances interact at the molecular level. In the case of the alkaline battery, the stoichiometric ratio between zinc (Zn) and manganese dioxide (MnO_{2}) from the equation is 1:2. This tells us:
  • 1 mole of Zn reacts with 2 moles of MnO_{2}
This ratio helps calculate the actual quantity of MnO2 involved based on the amount of Zn that reacts. Using this stoichiometric relationship, if 0.0688 moles of Zn are consumed, then the amount of MnO2 participating in the reaction is:\[\text{Moles of MnO}_{2} = 2 \times 0.0688 \, \text{mol} = 0.1376 \, \text{mol}\]Grasping stoichiometric ratios aids in predicting and measuring the quantities of substances needed or produced in chemical reactions.
Mole-to-Mass Conversion
Conversions from moles to mass use the concept of molar mass to translate the abstract idea of moles into measurable quantities of a substance. For MnO2, with a molar mass of 86.94 g/mol, converting the previously calculated moles to grams is straightforward.Use the equation:\[\text{Mass of MnO}_{2} = 0.1376 \, \text{mol} \times 86.94 \, \text{g/mol}\]to get 11.96 g of manganese dioxide.
  • Crucial for understanding the tangible amounts of substances in reactions.
  • Enables practical application of theoretical chemistry calculations.
Mastery of mole-to-mass conversions bridges the gap between theoretical chemistry and real-world applications.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode, or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

For the generic reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow\) \(\mathrm{A}^{-}(a q)+\mathrm{B}^{+}(a q)\) for which \(E^{\circ}\) is a positive number, answer the following questions: (a) What is being oxidized, and what is being reduced? (b) If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (c) Which half-reaction from (b) is higher in potential energy? (d) What is the sign of the free energy change for the reaction? [Sections \(20.4\) and \(20.5]\)

A common shorthand way to represent a voltaic cell is to list its components as follows: anode|anode solution || cathode solution|cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}\); sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+& 6 \mathrm{H}^{+}(a q)--\rightarrow \\ & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.

Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state. (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)-\cdots\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \rightarrow-\rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(a q)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

Two wires from a battery are tested with a piece of filter paper moistened with \(\mathrm{NaCl}\) solution containing phenolphthalein, an acid-base indicator that is colorless in acid and pink in base. When the wires touch the paper about an inch apart, the rightmost wire produces a pink coloration on the filter paper and the leftmost produces none. Which wire is connected to the positive terminal of the battery? Explain.
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