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Constants Fe2+ + 2e− = Fe(s) E º =−0.44 V; O + 4H+ + 4e− 2H2O...


Constants Fe2+ + 2e− = Fe(s) E º =−0.44 V; O + 4H+ + 4e− 2H2O E º =−1.23 V

2)In acid solution, the half cells Fe2+(aq) | Fe(s) and O2(g) | H2O are linked to create a spontaneous, galvanic cell.

a) Calculate Eºcell _____________
b) Which reaction occurs at the cathode _______________________________
c) Write the overall reaction for the cell ___________________________________
d) Write the shorthand notation for the anode (anode || cathode).

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Answer #1

Fe 2+ + 2 e− Fe(s)   E º =−0.44 V;

O 2 + 4 H+ + 4 e− 2 H2O E º =−1.23 V

Given reactions are reduction reactions as their is gain of electrons . The significance of electrode potential is that element having more positive value of reduction potential have more tendency to undergo reduction.

If we compare reduction potentials , it is clear that Fe will reduce and H2O will oxidise .  

The cell will be : H2O (l) I O 2(g) Il Fe 2+(aq) l Fe(s)

Reaction at anode :2 H2O (l)  O 2 (g) + 4 H+(aq) + 4 e−

Reaction at cathode :  2 Fe 2+(aq) + 4 e−2 Fe(s)

Overall reaction :   2 Fe 2+(aq) + 2 H2O (l)  O 2(g) + 4 H+(aq) + 2 Fe (s)

E 0cell = E 0cathode - E 0anode = -0.44 - (-1.23) =0.79 V

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