   # The total charge that ran he delivered by a large dry cell battery before its voltage drops too low is usually about 35 amp-hours. (One amp-hour is the charge that passes through a circuit when 1 A flows for 1 hour.) What mass of Zn is consumed when 35 amp-hours are drawn from the cell? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19, Problem 72GQ
Textbook Problem
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## The total charge that ran he delivered by a large dry cell battery before its voltage drops too low is usually about 35 amp-hours. (One amp-hour is the charge that passes through a circuit when 1 A flows for 1 hour.) What mass of Zn is consumed when 35 amp-hours are drawn from the cell?

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

The relation between solubility product Ksp and equilibrium constant is as follows.

Ksp= e+lnK

### Explanation of Solution

Given Amp = 35

Charge (C) = Current (A) × time (s)=35 A ×60 ×60 s=1.26×105 C

Let’s calculate the charge of the cell:

Charge(C) = (mole of e-)(96500 C1 mol e-)mol of e-=1.26×105 C96500 C=1

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