Question

The redox reaction of peroxydisulfate with iodide has been used for many years as part of the iodide clock reaction which introduces students to kinetics. If E not cell equals 1.587V and E not redox of the anode half cell is 0.536V, what is E not redox of the cathode half cell?. S2O82-(aq) + 2H+ +2I-(aq) yields 2HSO4-(aq) + I2 (aq). A. -1.051V. B. 1.051V. C. -2.123V. D. 2.123V. E. None of these choices is correct

Answer 1

We are given with a redox reaction described by the kinetics and the electrode potential of the anode half cell and cathode half cell. The electrode potential of the cell is the difference between the electrode potential of the cathode and the electrode potential of the anode. The equation is E cell = 1.587 V - 0.536 V equal to 1.051 V. Answer is B.

Answer 2

Answer : The correct option is, 2.123 V

Explanation : Given,

`E^o_(cell)=1.587V`

`E^o_(Anode)=0.536V`

The overall balanced equation of the cell is,

`S_2O_8^(2-)(aq)+2H^++2I^-\rightarrow 2HSO_4^-(aq)+I_2`

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:

`1.587V=E^o_(cathode)-0.536`

`E^o_(cathode)=2.123V`

Hence, the
`E^o` redox of cathode half cell is, 2.123 V