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  • A current of 3 amperes was passed for 2 hours through a solution of CuSO4. 3g of Cu2+ ions were discharged at the cathode. Calculate current efficiency. (Atomic mass of C

A current of 3 amperes was passed for 2 hours through a solution of CuSO4. 3g of Cu2+ ions were discharged at the cathode. Calculate current efficiency. (Atomic mass of Cu = 63.5)

(Response should be like 90.8 or 50.0 or 66.6)

Option: 1

62.4


Option: 2

42.2


Option: 3

36


Option: 4

92


Answers (1)

best_answer

Current Efficiency: It is the ratio of the mass of the products actually liberated at the electrode to the theoretical mass that could be obtained
\mathrm{C.E.}=\frac{\text { desired extent }}{\text { Theoretical extent of reaction }} \times 100 \%

 

\because \quad \mathrm{m}_{\mathrm{Cu}}\: =\: \mathrm{E \cdot i \cdot t / 96500}


\therefore \quad 3=\frac{63.5 \times i \times 2 \times 60 \times 60}{2 \times 96500}


So,          \mathrm{i=1.266 \text { ampere }}


\begin{array}{l}{\text { Current efficiency }} \\ {\qquad \begin{aligned} &=\frac{\text { Current passed actually }}{\text { Total current passed experimentally }} \times 100 \\\\ &=(1.266 / 3) \times 100=42.2 \% \end{aligned}}\end{array}

Posted by

Deependra Verma

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