The cell potential of the equation-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 the E0 half-cell value of euation-1 is 0.77 volts .
And value of E0 half cell of equation-2 is -0.44 volts .
By resolving or adding equation 1 and 2 we will get equation-3 ,
Thus , the value of the E0 half cell of equation-3 is given by ,
E0cell = 0.77-0.44 = 0.33 volts
Hence the E0 half cell of equation-3 is 0.33 volts .
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