Answer to Question #92041 in Physical Chemistry for Manula pasan

Question #92041
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.
1
Expert's answer
2019-07-29T04:30:13-0400

Solution.

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

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


(t+)=F×Δn(k)Q(t+) = \frac{F \times \Delta n(k)}{Q}Δw=w2w1\Delta w = w2-w1

Δw=0.194040.14941100=0.04463\Delta w = \frac{0.19404-0.14941}{100} = 0.04463

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

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

m(substance) = 0.0540 g


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

n(KCl) = 0.000724 mole


Δn(KCl)=0.000724mole\Delta n(KCl) = 0.000724 mole(t+)=96500×0.000724143.34=0.49(t+) = \frac{96500 \times 0.000724}{143.34} = 0.49

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

(t)=0.51(t-) = 0.51

(t+) = 0.49

(t-) = 0.51

Answer:

(t+) = 0.49

(t-) = 0.51


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