Advertisements
Advertisements
प्रश्न
Calculate the time required to deposit 2.4 g of \[\ce{Cu}\], when 2.03 A of current passed through \[\ce{CuSO4}\], solution. (At. mass of \[\ce{Cu}\] = 63.5 g mol−1)
उत्तर
Given:
Mass of \[\ce{Cu}\] deposited = 2.4 g
Molar mass of \[\ce{Cu}\] = 63.5 g mol−1
Current (I) = 2.03 A
Stoichiometry:
\[\ce{Cu^2+ + 2e- -> Cu}\]
`"Mole ratio" = (1 "mol Cu")/(2 "mol e"^-) = 1/2`
To find:
Time = t = ?
Formula:
`W = (I xx t)/96500 xx 1/2 xx 63.5`
`2.4 = (2.03 xx t)/96500 xx 1/2 xx 63.5`
`therefore t = (2.4 xx 96500 xx 2)/128.905`
`t = 463200/128.905`
= 3593.34 sec.
∴ The time required to deposit 2.4 g of \[\ce{Cu}\] is 3593.34 sec.
APPEARS IN
संबंधित प्रश्न
Answer the following in one or two sentences.
What is enthalpy of fusion?
Answer in brief.
What is the enthalpy of atomization? Give an example.
Define the Enthalpy of sublimation.
Define the Enthalpy of vaporization.
Why work done in vacuum is zero?
Define the Enthalpy of atomization.
The enthalpy change for the reaction; \[\ce{Na_{(s)} -> Na_{(l)}}\] is called enthalpy of ____________.
The enthalpy change for the transition of liquid water to steam, ∆vapH = 37.3 kJ mol−1 at 373 K. The entropy change for the process is ____________.
The enthalpy change for the chemical reaction \[\ce{H2O_{(s)} -> H2O_{(l)}}\] is called enthalpy of ______.
Define and explain enthalpy of freezing.
How does entropy change in the following process? Explain.
Condensation of vapour
Enthalpy of fusion of ice at 0°C and 1 atmosphere is 6.01 kJ mol - 1 and enthalpy of evaporation of water at 0°C and 1 atmosphere is 45.07 kJ mo1 - 1. Calculate the enthalpy of sublimation of one mole of ice at 0°C and 1 atmosphere. Write the equation for the same.
