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Chapter 20: Problem 74

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2} ?\)

Short Answer

Expert verified
The mass of MnO鈧� reduced at the cathode during dischange is approximately \(7.56 \ g\) and the number of electrical charge transferred from Zn to MnO鈧� is approximately \(131,700 \ C\).

Step by step solution

01

1. Balanced Redox Reaction

Determine the balanced redox chemical equation for the overall process occurring in the battery. The balanced redox chemical equation is given as: \[ Zn(s) + 2MnO_{2}(s) + 2H_{2}O(l) \rightarrow Zn(OH)_{2}(s) + 2Mn(OH)(s) \]
02

2. Stoichiometry for Mass of MnO鈧�

Calculate the stoichiometric relationship between Zn and MnO鈧� in the balanced equation. From the balanced equation, we can see that: 1 mole of Zn reacts with 2 moles of MnO鈧� Now, we can start calculating the mass of MnO鈧�.
03

3. Find Mass of MnO鈧�

Convert the given mass of Zn to moles and use the stoichiometry to find the mass of MnO鈧�. The molar mass of Zn = 65.38 g/mol First, convert the mass of Zn (4.50 g) to moles: \( moles \ of \ Zn = \frac{4.50 \ g}{65.38 \ g/mol} \) Then, use the stoichiometry from the balanced equation to find moles of MnO鈧� reduced: \( moles \ of \ MnO_{2} = 2 \times moles \ of \ Zn \) Finally, convert the moles of MnO鈧� to mass using its molar mass (MnO鈧� = 86.94 g/mol): \( mass \ of \ MnO_{2} = moles \ of \ MnO_{2} \times 86.94 \ g/mol \)
04

4. Coulombs of Electrical Charge

Determine the number of electrons transferred and use Faraday's constant to find the total coulombs. In the balanced reaction, 1 mole of Zn loses 2 moles of electrons. First, find the moles of electrons transferred: \( moles \ of \ electrons = 2 \times moles \ of \ Zn \) Then, use Faraday's constant (1 mol of electrons = 96,485 C) to calculate the coulombs: \( coulombs = moles \ of \ electrons \times 96,485 \ C/mol \) Now you can plug in the values calculated in the previous steps to find the mass of MnO鈧� and the number of coulombs transferred during the battery's discharge.

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Key Concepts

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

Stoichiometry
Stoichiometry is like a recipe in chemistry that tells us how much of each substance is needed or created in a chemical reaction. In our alkaline battery exercise, stoichiometry helps us determine how much manganese dioxide (\(\text{MnO}_{2}\)) reacts with zinc (\({Zn}\)).
In the balanced equation provided, each mole of zinc reacts with two moles of manganese dioxide. This is our stoichiometric relationship.
  • Molar mass of zinc is 65.38 g/mol. This lets us convert the given mass of zinc, 4.50 grams, into moles using the formula:
    \( moles \ of \ Zn = \frac{4.50 \ g}{65.38 \ g/mol} \)
  • Since 1 mole of Zn corresponds to 2 moles of \(\text{MnO}_{2}\), use this to find moles of \(\text{MnO}_{2}\)
  • Molar mass of \(\text{MnO}_{2}\) is 86.94 g/mol. Convert moles of \(\text{MnO}_{2}\) to grams using:
    \( mass \ of \ MnO_{2} = moles \ of \ MnO_{2} \times 86.94 \ g/mol \)
By following these steps, we harness stoichiometry to predict the precise amounts involved in our chemical process.
Coulomb's Law
Coulomb's Law in electrochemistry helps us understand the charge movement in reactions, especially within batteries. In our exercise, we need to calculate the electrical charge transferred during the reaction.
When zinc is oxidized, it loses electrons that then transfer to manganese dioxide, reducing it. This electron movement is an electric charge, measured in coulombs.
1 mole of zinc loses 2 moles of electrons, which is crucial for calculating transferred charge.
  • Convert moles of Zn (already found) into moles of electrons:
    \( moles \ of \ electrons = 2 \times moles \ of \ Zn \)
  • Use Faraday's constant, where 1 mole of electrons equals 96,485 coulombs:
    \( coulombs = moles \ of \ electrons \times 96,485 \ C/mol \)
Therefore, Coulomb's Law applied here gives us insight into the electrochemical charging process by linking electrons and coulombs.
Electrochemistry
Electrochemistry bridges chemistry and electricity, focusing on chemical reactions that produce or consume electric current. For our battery exercise, understanding electrochemistry is vital.
Batteries operate through redox reactions, where oxidation at the anode and reduction at the cathode occur.
  • Anode Reaction: Zinc (\(Zn\)) is oxidized, meaning it loses electrons and transforms into zinc hydroxide (\(Zn(OH)_2\)).
  • Cathode Reaction: Manganese dioxide (\(MnO_2\)) is reduced by gaining electrons, forming manganese hydroxide (\(Mn(OH)\)).
Electrochemistry provides us insights into the transfer of electrons and the resultant energy. This energy, harvested as electricity, powers devices. With thorough knowledge of electrochemistry, one can predict battery behaviors and optimize their efficiency.

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Most popular questions from this chapter

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (b) \(2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)

If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{-5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ}\).

In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging, approximately \(50 \%\) of the \(\mathrm{Li}^{+}\) ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO}_{2}\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

A voltaic cell utilizes the following reaction and operates at 298 K: $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M},\) \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M},\) and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M}\) and \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M} ?\)
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