A solution of CuSO 4 is electrolysed using a current of 1 ⋅ 5 amperes for 10 minutes. What mass of Cu is deposited at cathode ? (At. mass of Cu = 63 ⋅ 7 )

(D) The half reaction at cathode is Cu (aq) 2 + + 2 e −− ⟶ Cu ( s ) Mole ratio = 2 1 , t = 10 min = 600 s Mass of Cu deposited = 96500 I × t × mole ratio × molar mass of Cu = 96500 1.5 × 600 × 2 1 × 63.7 = 0.297 g

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minPYQ_2020

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A solution of $CuSO_{4}$ is electrolysed using a current of $1 \cdot 5$ amperes for 10 minutes. What mass of Cu is deposited at cathode ? (At. mass of $Cu = 63 \cdot 7$ )

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Answer:

Solution:

(D) The half reaction at cathode is $Cu_{(aq)}^{2 +} + 2 e^{--} ⟶ Cu_{(s)}$ Mole ratio $= \frac{1}{2}, t = 10 min = 600 s$ Mass of $Cu$ deposited $= \frac{I \times t}{96500} \times$ mole ratio $\times$ molar mass of $Cu$ $= \frac{1.5 \times 600}{96500} \times \frac{1}{2} \times 63.7 = 0.297 g$

Stream:MHT_CETSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2020

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