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Chapter 19: Problem 74

The cathode reaction in the Leclanché cell is given by $$ 2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s) $$ If a Leclanché cell produces a current of \(0.0050 \mathrm{~A}\) calculate how many hours this current supply will last if there are initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.

Short Answer

Expert verified
It will take approximately 493 hours for the cell to use up all of the \(MnO_{2}\) at a current of 0.0050 A.

Step by step solution

01

Calculating moles of MnO2 present

We first need to know how many moles of \(MnO_2\) are present. To calculate the moles of \(MnO_2\), we use the formula: Moles = Mass / Molar mass. For \(MnO_2\), the molar mass is approximately 86.94 g/mol. So, \(Moles_{MnO2} = \frac{4.0 g}{86.94 g/mol} = 0.046 mol\)
02

Calculating moles of electrons involved

Next, find the number of moles of electrons that will be involved in the reaction by using the stoichiometry of the reaction. This is done by multiplying the moles of \(MnO_2\) by 2 (since two electrons are involved per \(MnO_2\) molecule based on the chemical equation). So, \(Moles_{e-} = 2 \times Moles_{MnO2} = 2 \times 0.046 mol = 0.092 mol\)
03

Converting moles of electrons to charge

The moles of electrons can be converted into charge using Faraday's constant (1 Faraday = 96485 Coulombs/mol). This gives the total charge that will move through the circuit as the \(MnO_2\) is used up. So, \(Q = Moles_{e-} \times F = 0.092 mol \times 96485 C/mol = 8871 C\)
04

Calculating time taken for the reaction

Next, we calculate the time taken for this charge to flow using the relation \(Q = It\) where 'I' is the current and 't' is time. Solve for 't' to get \(t = \frac{Q}{I} = \frac{8871 C}{0.0050 A} = 1774200 s\)
05

Converting time from seconds to hours

Finally, to answer the question, we need to convert the time from seconds to hours. Knowing that 1 hour equals 3600 seconds, we get \(Time = \frac{1774200 s}{3600 s/hour} = 493 hours\)

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Most popular questions from this chapter

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