MHT CET · Chemistry · Electrochemistry
What is the charge required for the reduction of two moles of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) ?
- A \(2.89 \times 10^5 \mathrm{C}\)
- B \(1.93 \times 10^5 \mathrm{C}\)
- C \(9.65 \times 10^5 \mathrm{C}\)
- D \(3.86\times 10^5 \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(3.86\times 10^5 \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \)
\( 1 \mathrm{~mole} 2 \mathrm{~mole}^{-} \)
\( 2 \mathrm{~mole} 4 \mathrm{~mole}^{-} \)
\( \text {Charge of } 1 \mathrm{~mole} \text { electrons }=96500 \mathrm{C} \)
\( \text {Charge required for the reduction of two}\) \(\text{moles of } \mathrm{Cu}^{2+} \)
\( =4 \times 96500=3.86 \times 10^5 \mathrm{C}\)
\( 1 \mathrm{~mole} 2 \mathrm{~mole}^{-} \)
\( 2 \mathrm{~mole} 4 \mathrm{~mole}^{-} \)
\( \text {Charge of } 1 \mathrm{~mole} \text { electrons }=96500 \mathrm{C} \)
\( \text {Charge required for the reduction of two}\) \(\text{moles of } \mathrm{Cu}^{2+} \)
\( =4 \times 96500=3.86 \times 10^5 \mathrm{C}\)
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