Question

For the following electrochemical cell: Fe(s)| Fe3+(aq) || Cl(aq) | Cl2(g)|| Pt(s) a) Write the Cell Reaction for this Cell. b) Identify the Anode and Cathode.

Answer 1

SOLUTION

We have cell, Fe(s) I Fe³⁺(aq) II Cl^- (aq) I Cl₂(g) I Pt(s)

(a)

Anodic half cell reaction :     Fe(s) $\rightarrow$ Fe³⁺(aq) + 3e^-

Cathodic half cell reaction :    Cl₂(g) + 2e^-$\rightarrow$   2Cl^- (aq)

Adding two half reaction and balancing the reaction we will get the cell reaction

Multiplying the anodic half reaction by 2 and cathodic half reaction by 3 then adding the two reaction

2 Fe(s) +  3Cl₂(g) + 6e^-$\rightarrow$ 6Cl^- (aq) + 2Fe³⁺(aq) + 6e^-

Removing the common factor , we will get the cell reaction

2 Fe(s) +  3Cl₂(g) $\rightarrow$ 6Cl^- (aq) + 2Fe³⁺(aq) Answer

(b)

In writing the cell notation , first we write the anode then salt brige(ll) and then cathode

Fe(s) I Fe³⁺(aq) II Cl^- (aq) I Cl₂(g) I Pt(s)

Anode salt bridge Cathode

Anode :  Fe(s) I Fe³⁺(aq)

Anodic half cell reaction :     Fe(s) $\rightarrow$ Fe³⁺(aq) + 3e^-

Cathode :  Cl^- (aq) I Cl₂(g) I Pt(s)

Cathodic half cell reaction :    Cl₂(g) + 2e^-$\rightarrow$   2Cl^- (aq) Answer