NEET · Chemistry · STD 12 - 2. Electrochemistry
A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is :
(Given : Molar mass of \(Cu =63 g mol ^{-1} ; 1 F=96487 C mol ^{-1}\) )
- A 1.7018 g
- B 0.2938 g
- C 2.4036 g
- D 0.5876 g
Answer & Solution
Correct Answer
(B) 0.2938 g
Step-by-step Solution
Detailed explanation
(B) The reaction at the cathode is \(Cu ^{2+}+2 e ^{-} \rightarrow Cu\).
The quantity of charge \(Q\) passed through the solution is given by \(Q=I \times t\).
Substituting the given values:
\(Q=(A)5 A \times(10 \times 60) s=900 C\)
From Faraday's first law of electrolysis, the mass of copper deposited is:
\(m=\frac{M \times Q}{n \times F}\)
Here, \(M=63 g mol ^{-1}, n=2\), and \(F=96487 C mol ^{-1}\).
\(\begin{aligned} m=\frac{63 \times 900}{2 \times 96487}
m=\frac{56700}{192974} \approx 0.2938 g\end{aligned}\)
The quantity of charge \(Q\) passed through the solution is given by \(Q=I \times t\).
Substituting the given values:
\(Q=(A)5 A \times(10 \times 60) s=900 C\)
From Faraday's first law of electrolysis, the mass of copper deposited is:
\(m=\frac{M \times Q}{n \times F}\)
Here, \(M=63 g mol ^{-1}, n=2\), and \(F=96487 C mol ^{-1}\).
\(\begin{aligned} m=\frac{63 \times 900}{2 \times 96487}
m=\frac{56700}{192974} \approx 0.2938 g\end{aligned}\)
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