Question

A galvanic cell is based on the following half-reactions at 279 K: Ag+ + e- → Ag Eo = 0.803 V H2O2 (aq) + 2 H+ + 2 e- → 2 H2O Eo = 1.78 V What will the potential of this cell be when [Ag+] = 0.559 M, [H+] = 0.00393 M, and [H2O2] = 0.863 M?

Answer 1

Answer: The potential of the given cell is 0.856 V

Step-by-step explanation:

The substance having highest positive
`E^o` potential will always get reduced.

Half reactions for the given cell follows:

Oxidation half reaction:
`Ag\rightarrow Ag^(+)+e^-;E^o_(Ag^(+)/Ag)=0.803V` ( × 2)

Reduction half reaction:
`H_2O_2(aq.)+2H^(+)+2e^-\rightarrow 2H_2O;E^o=1.78V`

Net reaction:
`2Ag(s)+H_2O_2(aq.)+2H^(+)\rightarrow 2Ag^(+)+2H_2O`

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)=1.78-(0.803)=0.977V`

To calculate the EMF of the cell, we use the Nernst equation, which is:

`E_(cell)=E^o_(cell)-(2.303RT)/(nF)\log ([Ag^(+)]^2)/([H^(+)]^2[H_2O_2])`

where,

`E_(cell)` = electrode potential of the cell = ?V

`E^o_(cell)` = standard electrode potential of the cell = +0.977 V

R = Gas constant = 8.314 J/K mol

T = temperature = 279 K

F = Faraday's constant = 96500 C

n = number of electrons exchanged = 2

`[Ag^(+)]=0.559M`

`[H^(+)]=0.00393M`

`[H_2O_2]=0.863M`

Putting values in above equation, we get:

`E_(cell)=0.977-(2.303* 8.314* 279)/(2* 96500)* \log(((0.559)^2)/((0.00393)^2* (0.863)))\\\\E_(cell)=0.856V`

Hence, the potential of the given cell is 0.856 V