Question

In: Chemistry

What is the standard Gibb’s Free Energy of the reaction graphite | Br-(aq) | Br2(l) |...

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?

Solutions

Expert Solution

Oxidation half

2Br-(aq) ---> Br2(l) + 2e- Eo cell = 1.09 V

Reduction half

I2(s) + 2e---->2 I-(aq)      Eo 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

ΔGo = -nFEo

ΔGo = 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 = Eo - 0.0592 / n log ( Zn2+/ Ni2+)

Zn ---> Zn+2 +2e       Eo = 0.76 V

Ni2+ +2e -----> Ni(s) Eo = – 0.23 V

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

The value of n =2

Zn+2 [0.100 M)

Ni+2 (1.50 M)

E cell = Eo –( 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 = Eo –( 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


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