5. Suppose that you wish to attempt the separation of Cu from SbO by means of co

Question

5. Suppose that you wish to attempt the separation of Cu from SbO by means of contr pcaion of copper metal onto a platinuce e that the orl s- ple solution contains 1.00 MH,0.0311 MCu+, and 0.502 M Sbo, and that [H'] remains at 1.00 M throughout the course of the electrolysis. Cu2+ + 2 e- _ Cu ; Eo-+0.337 V vs. SHE SbO+ 2 H+ + 3 e- Sb + H2O ; Eo = +0.212 V vs. SHE (a) At what potential (versus SHE) must the platinum cathode be controlled to cause (b) What fraction of the Sbo* will be reduced to the elemental (Sb) state under the (c) If the volume of the original sample solution is 100.0 mL, what will be the total 999% deposition of elemental copper? experimental conditions imposed in part (a)? weight gain (in milligrams) of the platinum cathode under the experimental conditions imposed in part (a)?

Explanation / Answer

5a. As per Nernst equation:

E = E0Cu+2/Cu - (0.05916/n) log (1/[Cu+2]); where n = number of electrons involved and [Cu+2] is the conc. of Cu+2 at the end of the process.

Since we want 99.9% purity, 0.1% (i.e 0.001 fraction) will remain as Cu+2 after the elctrolysis.

therefore, [Cu+2] = 0.001* [Cu+2]o = .001*0.0311 M= 0.0000311M; Where, [Cu+2]o is the initial concetration of Cu+2.  ;

As per the question the Nernst equation comes down to : E = 0.337 - (0.05916/2)*log(1/0.0000311);

Solving this equation we get E = +0.204 V. Thus this is the required voltage.

2. Here we can apply the same Nernst equation. Here we know

E (i.e. +0.204V),

E0 = +0.212V

n = 3

[SbO+] = conc of SbO+ at the end which unknown and to be found

[H+] = 1M;

Replacing this values we get:

0.204 = 0.212 - (0.05916/3)*log(1/[SbO+][H+])

or, 0.008 = 0.0197*log(1/[SbO+]);

solving this equation we get [SbO+] = 0.392M

Therefore % of SbO+ converted = ([SbO+]initial - [SbO+]final) * 100/[SbO+]initial

= (0.512-0.392)*100/0.512 = 23.43%

c. The volume of sample = 100ml

Therefore amount of Cu deposited = Initial conc in molar * vol in L * Mol.wt. * 99.9/100

= 0.311*0.1* 65.54 * 99.9/100 =2.03gram

Amount of Sb deposited = 0.502 * 0.1* 137.76 * 23.43/100 = 1.62 g

Total weight increase = 2.03 + 1.62 = 3.65g = 3650 mg

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