Question

The following redox reaction occurs in a galvanic cell.2Al(s) + 3Fe²⁺ (1M) → 2Al³⁺ (1M) + 3Fe(a) Write the cell notation.(b) Identify anode and cathode.(c) Calculate E°cell if E°anod = -1.66V and E°cathode= -0.44V.(d) Calculate ΔG° for the reaction

Answer 1

a)  $Al/Al^{3+}(1M)//Fe^{2+}(1M)/Fe$

b) Aluminium : anode.

Iron: cathode.

c) $E^0=-0.44- (-1.66V)=1.22V$

d) $\Delta G=-2\times 96500\times 1.22=-706380J$

Explanation:

$2Al(s)+3Fe^{2+}\rightarrow 2Al^{3+}+3Fe$

a) The cell notation is $Al/Al^{3+}(1M)//Fe^{2+}(1M)/Fe$

b) Here Aluminium undergoes oxidation by loss of electrons, thus act as anode.Iron undergoes reduction by gain of electrons and thus act as cathode.

c) $E^0=E^0_{cathode}-E^0_{anode}$

Where both are standard reduction potentials.

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=-0.44- (-1.66V)=1.22V$

d) The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

= gibbs free energy

n= no of electrons gained or lost

F= faraday's constant

$E^0$ = standard emf

$\Delta G=-2\times 96500\times 1.22=-706380J$

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