Discussion Forum

1)

For the electrochemical cell,

                Mg(s) | Mg ²⁺ (aq. 1M)|

                                           |Cu²⁺ (aq. 1M)| Cu(s)

 The standard emf of the cell is 2.70V at 300K. When the concentration of Mg ²⁺ is changed to x M, the cell potential changes to 2.67 V at 300K. The  value of x is .............. [ Given, $\frac{F}{R}=11500KV^{-1}$, where F is the Faraday constant and R is the gas contant, ln (10)=2.30)

Answer:

Option A

Explanation:

Equation  of cell reaction according  to the cell notation given, is

Given $E_{cell}^0=2.7V$. T= 300K

with [ Mg ²⁺ (aq)]= 1 M and [Cu ²⁺ (aq)]= 1 M and n=2

  Further . E_cell =2.67 V

  with [Cu ^2- (aq)]= 1M and [Mg ²⁺ (aq)]=xM

    and  $\frac{F}{R}=11500KV^{-1}$

   where F= Faraday constant,  R= gas constant

From the formula,

     $E_{cell}=E_{cell}^{0}-\frac{RT}{nF}ln\frac{[Mg^{2+}(aq)]}{[Cu^{2+}(aq)]}$

   After putting the given values

      $2.67= 2.70- \frac{RT}{2F}ln\frac{x}{1}$

or     $2.67= 2.70- \frac{RT}{2F}ln x$

           $-0.03=  \frac{-R\times 300}{2F}lnx$

Or      $ln x= \frac{0.003\times 2}{300}\times\frac{F}{R}$

             = $\frac{0.003\times 2 \times11500}{300}=2.30$

   So, ln x= 2.30

  or x=10 (as given  ln(10)=2.30)