The E°half-cell of the (3) is 0.33 volts .
Given,
(1) Fe3+(aq) +e =Fe2+(aq)
(2) Fe2+(aq) +2e = Fe(s)
(3) Fe3+(aq) +3e = Fe(s)
We know ,
E° half-cell of (1) is 0.77 volts .
E° half cell of ( 2) is -0.44 volts .
By resolving or adding equation 1 and 2 we will get (3) ,
Thus , the value of the E° half cell of (3) is given by ,
E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts
Hence the E° half-cell of ( 3 ) is 0.33 volts .
<h3>What is cell potential? </h3>
It is the difference between the electrode potentials of two half cells.
it also defined as the force which causes the flow of electrons from one electrode to the another and thus results in the flow of current.
Ecell = E( cathode) - E ( anode)
<h3>What is electrode potential ? </h3>
The tendency of the electrode to lose of gain electrons is called as electrode potential.
Learn more about cell potential here :
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