Question

If the standard potential for Daniel cell is 1.1 V, then the potential of the cell when [Zn 2+] = 1.0M and [ Cu 2+ ] =0.1 M at 298K is (2 .303 RT/F value at 298K is0.06 V)

Answer 1

Potential of cell is 1.07 V

Explanation:

In Daniel cell, Zn is oxidised at anode and Cu is reduced at Cathode.

$Zn_{(s)}\rightarrow Zn_{(aq)}^{+2}+2e^{-}$

$Cu_{(aq)}^{+2}+2e^{-}\rightarrow Cu_{(s)}$

Overall reaction is,

$Zn_{(s)}+Cu_{(aq)}^{+2}\rightarrow Zn_{(aq)}^{+2}+Cu_{(s)}$

So, the Nernst equation for Daniel cell is,

$E_{cell}=E^{0}_{cell}-\frac{2.303RT}{nF}log\frac{[Cu_{(s)}][Zn^{+2}_{(aq)}]}{[Zn_{(s)}][Cu_{(aq)}^{+2}]}$

For Pure solid molar concentration or active mass is always 1.

$E_{cell}=E^{0}_{cell}-\frac{2.303RT}{nF}log\frac{[Zn^{+2}_{(aq)}]}{[Cu_{(aq)}^{+2}]}$

Given $E^{0}_{cell}=1.1V, \frac{2.303RT}{F}=0.06V$ , n =2

$[Zn^{+2}]$ = 1.0 M , $[Cu^{+2}]$ = 0.1 M

Put these values in above equation

$E_{cell}=1.1-\frac{0.06}{2}log\frac{1}{0.1}$

$E_{cell}=1.1-\frac{0.06}{2}log10$   (log 10= 1 )

$E_{cell}=1.1-\frac{0.06}{2}\times 1$

$E_{cell}=1.1-0.03$

$E_{cell}=1.07V$

So, the potential of cell is 1.07 V.