ElectrochemistryHard
Question
Copper sulphate solution (250 ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 19.3 min. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.
Options
A.9.6 × 10−5 M
B.4.8 × 10−5 M
C.2.4 × 10−5 M
D.1.2 × 10−5 M
Solution
$n_{eq}Cu^{2 +}\text{ reduced = }\frac{Q}{F} \Rightarrow n \times 2 = \frac{2 \times 10^{- 3} \times 19.3 \times 60}{96500}$
$\therefore n = 1.2 \times 10^{- 5} $$$\therefore\left\lbrack CuSO_{4} \right\rbrack_{0} = \frac{\left( 1.2 \times 10^{- 5} \times 2 \right)}{250} \times 1000 = 9.6 \times 10^{- 5}\text{ M}$$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
The cell Zn|Zn²⁺(1M)||Cu²⁺(1M)|Cu with $E^\circ_{\text{cell}} = 1.10$ V was allowed to be completely discharged at 298 K...In a conductivity cell, the two platinum electrodes, each of area 10 cm2 are fixed 1.5 cm apart. The cell contained 0.05...Using the data in the preceding problem, calculate the equilibrium constant of the reaction at 25oC.Zn + Cu++ ⇋ Zn...A lead storage battery has initially 200 g of lead and 200 g of PbO2 plus excess H2SO4. Theoretically, how long could th...The standard potentials of MnO4−|Mn2+ and MnO2|Mn2+ electrodes in acid solution are 1.51 and 1.23 V, respectively. Stand...