Question

A sample of 1500 mg of an alloy that contains silver, copper, and chromium is dissolved and the solution containing Ag+ , Cu2⁺, and Cr³⁺ ions, is diluted to exactly 500 cm³. One tenth of the volume of that solution is taken for further procedure: After elimination of silver and copper, chromium is oxidised in it according to the following unbalanced equation: 

OH^- + Cr³⁺ H₂O₂ CrO₂^-4 +H₂O

Then 25.00 cm³ of a 0.100 molar Fe(II) salt solution are added. The following reaction (written in an unbalanced form) is taking place: 

H⁺ Fe²⁺ CrO^2-₄ Fe³⁺ Cr³⁺ H₂O

According to the unbalanced equation: 

H⁺ +Fe²⁺ +MnO^-₄ Fe³⁺ +Mn²⁺ +H₂O

a volume of 17.20 cm³ of a 0.020-molar KMnO⁴ solution is required for an oxidation of the Fe(II) salt which remains unoxidized in the solution. In another experiment, a volume of 200 cm³ of the initial solution is electrolysed. Due to secondary reactions, the efficiency of the electrolysis is 90 % for metals under consideration. All three metals are quantitatively deposited in 14.50 minutes by passing a current of 2 A through the solution.

 Balance the three chemical equations and calculate the composition of the alloy in % by mass. 

Relative atomic masses: A_r(Cu) = 63.55; A_r(Ag) = 107.87; A_r(Cr) = 52.00

Answer 1

Reacted: 25 × 0.1 − 1.72 = 0.78 mmol Fe²⁺  

It corresponds: 

0.78/3 = 0.26 mmol Cr in 150 mg of the alloy

m(Cr) = 2.6 mmol × 52 g mol^-1 = 135.2 mg in 1500 mg of the alloy % Cr = 9.013

Content of Cu and Ag: 

Q = 40.575 mF / 1500 mg (1087.4 mAh) 

QCr = 2.6 × 3 = 7.8 mF (209 mAh) 

Q(Cu+Ag) = 40.575 − 7.8 = 32.775 mF (878.4 mAh) 

(F = Faraday's charge) 

m(Cu + Ag) = m(alloy) − m(Cr) = 1500 − 135.2 = 1364.8 mg

For deposition of copper:  2x/63 .55 mF

For deposition of silver: 1364 8. − x/107. 87 mF

32.775 =2x/63.55 +1364.48-x/107.87 mF

x = 906.26 

m(Cu) = 906.26 mg in 1500 mg of the alloy 

m(Ag) = 458.54 mg in 1500 mg of the alloy  

% Cu = 60.4 % Ag = 30.6