Question

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2+(aq)+2e−→Cu(s) and Fe(s)→Fe2+(aq)+2e− The net reaction is Cu2+(aq)+Fe(s)→Cu(s)+Fe2+(aq) Use the given standard reduction potentials in your calculation as appropriate.

Answer 1

The standard cell potential of the reaction is 0.78 Volts.

Step-by-step explanation:

`Cu^(2+)(aq)+Fe(s)\rightarrow Cu(s)+Fe^(2+)(aq)`

Reduction at cathode :

`Cu^+(aq)+2e^-\rightarrow Cu(s)`

Reduction potential of
`Cu^(2+)` to Cu=
`E^o_(1)=0.34 V`

Oxidation at anode:

`Fe(s)\rightarrow Fe^(2+)(aq)+2e^-`

Reduction potential of
`Fe^(2+)` to Fe=
`E^o_(2)=-0.44 V`

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

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

Putting values in above equation, we get:

`E^o_(cell)=0.34V -(-0.44 V)=0.78 V`

The standard cell potential of the reaction is 0.78 Volts.