Calculate the concentrations needed to make 1.00V and 0.80V galvanic cell with c

Question

Calculate the concentrations needed to make 1.00V and 0.80V galvanic cell with cobalt (II) and manganese (II), if the more concentrated metal ion can be prepared with a concentration of 0.20M. Clearly indicate the concentration for each ion present in each cell. 4. For each skeletal chemical equation, create the simplest balanced chemical equation. In acid solutions: Cu(s) + NO_3^- (aq) rightarrow Cu^2+ (aq) + NO(g) Al(s) + MnO_4^- (aq) rightarrow Al(OH)_4 (aq) + MnO_2(s) Sn^2+ (aq) + Pb(s) + SO_4^2- (aq) Sn(s) + PbSO_4(s) In basic solution: Br_2(I) rightarrow BrO_3^- (aq) + Br^- (aq) NO_2^- (aq) + Al(s) rightarrow NH_3 (g) + AlO_2^- (aq) 5. Calculate the E_cell for the last reaction shown above, assuming 8degree C and with [Sn^2+] = 0.060M, and [SO_4^2+] = 0.040M. 6. A certain electrolytic cell is used to make Mg from Mg^2+ ions, using a current of 7.5amperes. How long would the cell have to operate to produce 6.0mol Mg?

Explanation / Answer

6

Explanation & Calculations :.

(i) The question is based upon Faraday's laws of electrolysis. Accordingly the following relation is used to calculate the time ( in seconds ) required to deposited the given mass ( = number of moles x atomic mass ) of Magnesium from Mg2+  (aq ) solution.

.....................w = z c t

where , w represents the weight of metal Mg

.....................................................(given in terms of moles = 6.00 moles )

............................................................................................=( 6.00 x 24.31 ) gms.

..................c denotes current in amperes ....... (given as = 7.5 amp . )

..................z is Elecrochemical Equivalent weight * of the metal being deposited

................ for magnesium it is = [ ( 24.31 / 2 ) / 96 .485.3399 ] coulomb / g

..................................................= ( 1.25563 x 10-4 ) coulomb / g

*[ ECE ( electrochemical equivalent weight represents the quantity of electricity ( in coulombs ) required to deposit one gram equivalent weight of a metal .]

..............t represents time in seconds .......= ? ( tobe calculated )

(ii) Substituting the given values in the expression we get ,

................6 .00 x 24.31 = 1.25563 x 10-4 x 7.5 x t   

...............................or , t = ( 6.00 x 24.31 ) / 1.25563 x 10-4  x 7.5

.........................................= 154886.39 seconds

..................................or,...= (154886.39 / 3600 ) hours

.........................................= 46.02 hours

Thus the cell has to be operated for 46.02 hours to produce 6.0 mols of Mg

____________________________________________________________________________

5.

Calculations

(i) Calculate Kc  for the given reaction & then apply the relation

.......................Ecell   = 2.303RT / n F (log Kc ) }

(ii).............Kc   ={ [ PbSO4 (s) ] [ Sn (s) ] / [ Sn2+ ] [ SO42- ] [ Pb(s) ] }

.....................={ ( 1.0 x 1.0 ) / ( 0.060 x 0.040 x 1.0 ) }

....................= 2.40 x 10-3

( note that concentrations of solids in pure state is constant and is taken as unity )

(iii)

....................Ecell =( 2.303 x 8.3145 x 281 / 2 x 96485.3399 ) log (2.40 x 10-3 )

..............................= 0.02788 log (2.40 x 10-3 ) V

..............................= - 0.073 V , ..........at 8o C

where , T = ( 273 + 8 ) = 281 K , R = 8.3145 J / mole K , F = 96485.3399 coulombs , n = 2 , Kc  = 2.40 x 10-3

______________________________________________________________________

Please submit the other two questions separately as fresh questions , glad to help.

  

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.