Question

The following redox reaction occurs in a galvanic cell.

2Al_(s) + 3Fe²⁺ (1M) → 2Al³⁺ (1M)+ 3Fe_(s)

(a) Write the cell notation. 

(b) Identify anode and cathode

(c) Calculate E⁰_Cell if E⁰_anode = – 1.66 V and E⁰_cathode  = – 0.44 V 

(d) Calculate ΔG⁰ for the reaction.

Answer 1

Given :

E⁰_anode = \(E^0_{Al^{3+}/Al}\) = 1.66V;

Overall cell reaction :

2Al_(s) + 3Fe²⁺ (1M) → 2Al³⁺ (1M)+ 3Fe_(s)

(a) In the cell reaction, 

Al is oxidised from zero to 3+ while Fe³⁺ is reduced from 3+ to zero.

Hence the cell notation is,

(b) Anode : Al electrode at LHE 

Cathode : Fe electrode at RHE

(d) The standard free energy change ΔG⁰ is given by,

ΔG⁰ = – nFE⁰_cell

= – 6 × 96500 × 1.22 

= – 70640 J 

= – 706.4 kJ

∴ (a) Cell notation :

Al_(s) | \(Al^{3+}_{(aq)}\)(1M) || \(Fe^{2+}_{(aq)}\)(1M)|Fe_(s)

(b) Anode : Al; 

Cathode : Fe

(c) E⁰_cell = 1.22 V

(d) ΔG⁰ = – 706.4 kJ