Question

0.14941 wt% aqueous KCl solution
at 25°c was electrolyzed in a hittorf apparatus using two Ag-AgCl electrods .The cathode reaction was AgCl+ e ~ Ag + Cl- , the anode reaction was the reverse of this .After the experiment ,it was found that 143.34 C charge had been flown through the apparatus and that the cathode compartment contained 120.99g of solution
that was 0.19404wt% KCl by weight.Calculate t+ and t- in the KCl solution
used in the experiment ,neglect the transport of water by ions.

Answer 1

Solution.

$$(t+) + (t-) = 1$$

To determine the number of transfer of potassium ions and chlorine, we write the following formula:

$$(t+) = \frac{F \times \Delta n(k)}{Q}$$ $$\Delta w = w2-w1$$

$$\Delta w = \frac{0.19404-0.14941}{100} = 0.04463$$

$$\Delta w = \frac{m(substance) \times 100}{m(solution)}$$

$$m(substance) = \frac{\Delta w \times m(solution)}{100}$$

m(substance) = 0.0540 g

$$n(KCl) = \frac{m}{M}$$

n(KCl) = 0.000724 mole

$$\Delta n(KCl) = 0.000724 mole$$ $$(t+) = \frac{96500 \times 0.000724}{143.34} = 0.49$$

$$(t-) = 1 - (t+)$$

$$(t-) = 0.51$$

(t+) = 0.49

(t-) = 0.51

Answer:

(t+) = 0.49

(t-) = 0.51