Chemistry Electrochemistry

A salt bridge is a U-shaped tube that connects the two halves (oxidation and reduction) of the cell with an electrochemical cell, such as a galvanic cell. It allows the flow of ions in both halves and maintains the electricity to stay neutral.

In electrochemistry, metallic and electrolytic are two types of conductors.

The two branches of electrochemistry are: Electricity generated by chemical reactions and Chemical reactions that generate electricity.

Eq. = no. of faraday

$\frac{10*0.2*10^{-2}*19.7}{197}*3=\frac{0.193*t}{96500}$

$G\frac{\mathcal{l}}{a}=\kappa$ .1

${\Lambda }_m=\kappa *\frac{1000}{n/\left(\mathcal{l}*a\right)*10^{-3}}$ .2

from (1) & (2) ${\Lambda }_m=2{\Omega }^{-1}-c{m}^2-mo{l}^{-1}$

ion flow only   

Charge (q)   $=it=2*50*60=6000\mathrm{C}$

$\Rightarrow \mathrm{}6000C=\left(\frac{6000}{96500}F\right)$

${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}+3{\mathrm{e}}^{(-)}+2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{weakly alkaline medium}}{?}{\mathrm{M}\mathrm{n}\mathrm{O}}_2+4{\mathrm{O}\mathrm{H}}^{-}$

Mass of ${\mathrm{M}\mathrm{n}\mathrm{O}}_2$    deposited $=\frac{1}{3}*\frac{6000}{96500}*87=1.8$

Mol wt of propane = 44 g and weight of propane = 12.0 g
So, using PV = nRT
V = (nRT)/P = (12/44) * 0.082 * 333) / (740/760) = 7.648 L

$AgBr?A{g}^{+}+B{r}^{-}$                

S + 10^-7                        S

(S + 10^-7) S = 12 * 10^-14

  $\therefore S=3*10^{-7}M$          

${\kappa }_{totel}=\left[6*4*10^{-7}\right]+\left[8*3*10^{-7}\right]+\left[7*1*10^{-7}\right]$          

= 55 * 10^-7 S m^-1

Anode      Ag + Br^- -> AgBr + e^-

Cathode   Ag⁺ + e^- ® Ag

  ${E^o}_{cell}={\left({E^o}_{RP}\right)}_C-{\left({E^o}_{RP}\right)}_A$        

$\begin{array}{l}{E}_{cell}=-0.06log{K}_{sp}-0.06log100\\ 0.643=-0.06log{K}_{sp}-0.12\\ log{K}_{sp}=-\frac{0.763}{0.06}=-12.71\end{array}$