Question
Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks!
Given the voltaic cell notation, Cu(s) lCu+2 ll Fe+3 l Fe+2,calculate: (a) the equilibrium constant for the cell netreaction. (b) the actual work obtained when from 0.112 kg ofiron that reactions.
Given the voltaic cell notation, Cu(s) lCu+2 ll Fe+3 l Fe+2,calculate: (a) the equilibrium constant for the cell netreaction. (b) the actual work obtained when from 0.112 kg ofiron that reactions. (b) the actual work obtained when from 0.112 kg ofiron that reactions.
Explanation / Answer
The given voltaic cell notation is Cu(s) lCu+2 ll Fe+3 l Fe+2 First we Write the equationsfor the cell half-reactions, calculate the standard cellpotentialand determine the number of electrons transferred. Cu (s)-------> Cu2+ +2e- E0Oxidation = -0.337 V 2Fe+3 +2e- -------- >2Fe+2 E0reduction = 0.700 V ----------------------------------------------------------------------------------------- Cu(s)+2Fe3+(aq)------> Cu2+(aq)+2 Fe2+ E0cell = +0.363V n = 2 moles From the ractionE0cell = (0.0592/n) log K +0.363V=( 0.0592 / 2 )logK = 0.0296 logK log K= 0.363 / 0.0296 =12.26 K = 1012.26 K= 1.8*1012 The total work obtained from the cell isG =- nFE0 =-2 * 96,500 J/V* 0.363 V =-70059 J K= 1.8*1012 The total work obtained from the cell isG =- nFE0 =-2 * 96,500 J/V* 0.363 V =-70059 J