Question

Enter electrons as e. Use smallest possible integer coefficients for ALL reactions. If a box is not needed, leave it blank. A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt bridge Cl2(g)+2Fe (aq) +2Cr(aq)+ 2Fe (aq) The anode reaction is: The cathode reaction is: the Fe Fe electrode the CI Cl2 electrode. In the external circuit, electrons migrate In the salt bridge, anions migrate [ the Fe Fe compartment the CI Cl, compartment.

Answer 1

1.

Anode reaction:

Fe²⁺ - e $\to$ Fe³⁺   (1)

Cathode reaction

2 Cl^- + 2 e $\small \to$ Cl₂ (2)

Spontaneous cell reaction

Eq.1 * 2 + Eq. 2

2 Fe²⁺ (aq) + 2 Cl^- (aq)  $\small \to$ 2Fe³⁺ (aq) + Cl₂ (g)

On external circuit electrons migrate from anode Fe²⁺ | Fe³⁺ electrodeto cathode Cl^- | Cl₂ electrode.

in saltbridge anions migrate to Fe²⁺ | Fe³⁺ compartment from  Cl^- | Cl₂​​​​​​​ electrode.

2.

Anode reaction

Mn - 2e $\small \to$ Mn²⁺ ; E⁰red ( Mn²⁺/Mn) = - 1.180 V  

Cathode reaction

Pb²⁺ + 2e $\small \to$ Pb ;  E⁰_{red(Pb²⁺/Pb)}  = - 0.126 V

spontaneous cell reaction

Mn (s) + Pb²⁺ (aq)  $\small \to$ Mn²⁺ (aq) + Pb (s)

cell potential ; E^o_cell = E⁰_{red ( cathode)} -E⁰_{red( anode)} = - 0.126 - ( -1.180) = 1.054 V

3.

Anode reaction

Hg - 2e $\small \to$ Hg²⁺ ; E⁰_{red ( Hg²⁺/Hg)} = 0.855 V

Cathode reaction

2 Cl^- + 2 e $\small \to$ Cl₂ (2)   E⁰_red(Cl^-/Cl)  = 1.360 V

spontaneous cell reaction

Hg (l)) + 2 Cl ^- (aq)   $\small \to$ Hg²⁺ (aq) + Cl₂(s)

cell potential ; E^o_cell = E⁰_{red ( cathode)} - E⁰_{red( anode)} = 1.360 - ( 0.855) = 0.505 V