Question

Calculate the emf of the cell reaction at 25°C:

\[\ce{Zn_{(s)}_{(0.1 M)} {|} Zn^{2+}_{ (aq)}_{(0.1 M)} {||} Cd^{2+}_{ (aq)}_{(0.01 M)} {|} Cd_{(s)}}\]

Given: `E_(Cd^(2+)//Cd)^° = −0.40  V`

`E_(Zn^(2+)//Zn)^°= −0.76  V`

[log 10 = 1]

Answer 1

Given: The given electrochemical cell is:

\[\ce{Zn_{(s)}_{(0.1 M)} {|} Zn^{2+}_{ (aq)}_{(0.1 M)} {||} Cd^{2+}_{ (aq)}_{(0.01 M)} {|} Cd_{(s)}}\]

`E_(Zn^(2+)//Zn)^° = −0.76  V`

`E_(Cd^(2+)//Cd)^° = −0.40  V`

Standard EMF of the cell:

`E_"cell"^° = E_"cathode"^ ° - E_"anode"^ °`

= (−0.40) − (−0.76)

= −0.40 + 0.76

= 0.36

The cell reactions are:

\[\ce{Zn_{(s)} -> Zn{^{2+}_{(aq)}} + 2e-}\] (At anode)

\[\ce{Cd^{2+}_{ (aq)} + 2e- -> Cd_{(s)}}\] (At cathode)

\[\ce{Zn_{(s)} + Cd^{2+}_{ (aq)}-> Zn^{2+}_{ (aq)} + Cd_{(s)}}\] (Cell reaction)

n = 2

Nernst equation for the cell reaction is:

= `E_"cell"^° - 0.0591/n log  ([Zn^(2+)])/([Cd^(2+)])`

= `0.36 - 0.0591/2 log  0.1/0.01`

= `0.36 - 0.0591/2  log  10`

= `0.36 - 0.0591/2 xx 1`

= 0.36 − 0.02955 × 1

= 0.330 V