Question

What is the standard Gibb’s Free Energy of the reaction graphite | Br^-(aq) | Br₂(l) | | I₂(s) | I^-(aq)? Is the reaction spontaneous under standard conditions?

What is the initial cell potential of the voltaic cell represented by Zn | Zn⁺² [0.100 M) | | Ni⁺² (1.50 M) | Ni ?

b. What is the cell potential when [Ni⁺]=0.500 M?

c. What are the metal ion concentrations when the cell potential has fallen to 0.45 V?

Answer 1

Oxidation half

2Br-(aq) ---> Br₂(l) + 2e- E^o cell = 1.09 V

Reduction half

I₂(s) + 2e---->2 I-(aq)      E^o cell = - 0.54 V

2I- (aq)+ Br₂ (aq) ----> I₂ (s) + 2 Br- (aq)

E⁰cell = 1.09 V -0.54 V = +0.55 V

For the reaction

ΔG^o = -nFE^o

ΔG^o = standard Gibb’s Free Energy of the reaction  

n = 2 is the number of moles of electrons transferred

F is Faraday’s constant 1F = 96,500 C/mol = 96,500 J/Vmol

= -2 /mol*96,485 J/V*0.55 V= - 106133.5 J/ mol = - 106.13 kJ/mol

ΔG is negative, so this reaction is spontaneous.

E cell = E^o - 0.0592 / n log ( Zn²⁺/ Ni²⁺)

Zn ---> Zn⁺² +2e       E^o = 0.76 V

Ni²⁺ +2e -----> Ni(s) E^o = – 0.23 V

Zn(s) + Ni²⁺(aq)--->Zn²⁺(aq) + Cu(s) E^o = 0. 99 V

The value of n =2

Zn+2 [0.100 M)

Ni+2 (1.50 M)

E cell = E^o –( 0.0592 / n ) log ( Zn²⁺/ Ni²⁺)

          = 0.99 V – (0.0592 / 2 ) log (0.100 /1.50)

          = 0’99 V - 0.0296 * - 1.176

          = 0.99 V – (-0.035) = 1.025 V

b. The cell potential when [Ni⁺]=0.500 M

E cell = E^o –( 0.0592 / n ) log ( Zn²⁺/ Ni²⁺)

          = 0.99 V – (0.0592 / 2 ) log (0.100 /0.50)

          = 0.99 V - 0.0296 * - 0.699

         = 0.99 V –(- 0.0207) =1.010 V