A galvanic cell whose cell reaction is 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq) has a cell potential of 0.72V. What is the maximum electrical work that can be obtained from this cell per mole of iron(III) ion?

138.96kJ is the maximum electrical work

Explanation:

The maximum electrical work that can be obtained from a cell is obtained from the equation:

W = -nFE

Where W is work in Joules,

n are moles of electrons = 2mol e- because half-reaction of Zn is:

Zn(s) → Zn²⁺(aq) + 2e⁻

F is faraday constant = 96500Coulombs/mol

E is cell potential = 0.72V

Replacing:

W = -2mol*96500Coulombs/mol*0.72V

W = - 138960J =

the answer is either b or c

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.