Chemistry Electrochemistry

In simple words, the study of relationship between electrical energy and chemical reactions is called the Electrochemistry. The concept comprises how chemical reactions can create electrical energy and how electrical energy can generate chemical changes.

$C{a}_3{\left(P{O}_4\right)}_2\to 3C{a}^{2+}+2P{O}_4^{3-}$

$2Cr+3F{e}^{2+}\stackrel{n=6}{\to }2C{r}^{3+}+3Fe$

$?G^0=-n{E}^{0}F$

$170*{10}^3=-6*E^0*96500$

$E_{cell}^0=0.29V$

Order of limiting molar conductivity

$O{H}^{-}>S{O}_4^{2-}>C{l}^{-}>C{H}_{3}CO{O}^{-}$

More the number of ions in solution, more will be the specific conductance

 Ionic conductivity of H⁺ is more than that of Na+

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :  $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$    

Cathode :  2Ag⁺(aq) + 2e^- -> 2Ag(s)

Zn(s) + 2Ag⁺(aq) -> Zn²⁺ (aq) + 2Ag(s)

  $E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$          

x = 147

H₂ (g) + Cu²⁺ (aq) $\stackrel{}{\to }$  2H⁺ (aq) + Cu (s)

  $\therefore E_{cell}=E_{cell}^0-\frac{2.303RT}{nf}logQ$

0.31 = 0.34 - $\frac{0.06}{2}log\frac{\left[H^{+}\right]}{C{u}^{2+}}$  

  $\left[C{u}^{2+}\right]=10^{-7}\left(?\left[H^{+}\right]=10^{-3}\right)$

x = 7

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

Using ; k_sp = 108 S⁵

1.1 * 10^-23 = 108 S⁵

S =  ${\left(\frac{110}{108}*10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5*1000}S{m}^{2}mo{l}^{-1}$  

= 3 * 10^-3 Sm² mol^-1

So; x = 3

H₂ (g) + Cu²⁺ (aq) $\stackrel{}{\to }$  2H⁺ (aq) + Cu (s)

$\therefore E_{cell}=E_{cell}^0-\frac{2.303RT}{nf}logQ$  

0.31 = 0.34 -  $\frac{0.06}{2}log\frac{\left[H^{+}\right]}{C{u}^{2+}}$  

$\left[C{u}^{2+}\right]=10^{-7}\left(?\left[H^{+}\right]=10^{-3}\right)$  

x = 7