Question

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)where:ClO2+e-→ClO2-Ered0 =+0.954 V.Cl2+2e-→2Cl-Ered0 =+1.36 V.a. ΔG=+79 kJb. ΔG=-790 kJc. ΔG=-79 kJd. ΔG=-0.79 kJ

Answer 1

Answer: The
`\Delta G^o` for the given reaction is
`-7.84* 10^4J`

Step-by-step explanation:

For the given chemical reaction:

`2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)`

Half reactions for the given cell follows:

Oxidation half reaction:
`ClO_2^-\rightarrow ClO_2+e^-;E^o_(ClO_2^-/ClO_2)=0.954V` ( × 2)

Reduction half reaction:
`Cl_2+2e^-\rightarrow 2Cl(g);E^o_(Cl_2/2Cl^-)=1.36V`

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(cathode)-E^o_(anode)`

Putting values in above equation, we get:

`E^o_(cell)=1.36-(0.954)=0.406V`

To calculate standard Gibbs free energy, we use the equation:

`\Delta G^o=-nFE^o_(cell)`

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

`E^o_(cell)` = standard cell potential = 0.406 V

Putting values in above equation, we get:

`\Delta G^o=-2* 96500* 0.406=-78358J=-7.84* 10^4J`

Hence, the
`\Delta G^o` for the given reaction is
`-7.84* 10^4J`