Answer:
10.035 g of Cu deposited.
Explanation:
Equation of the reaction:
Cu(NO3)2 --> Cu2+ + 2NO3^-
Q = It
Where,
I = current
= 3.53 A
t = time
= 1.2 h * 3600 s/1 h
= 4320 s
Q = 4320 * 3.52
= 15249.6 C
1 mole of metal deposited contains 96500C.
Therefore,
Number of moles = 15249.6/96500
= 0.158 mol of Cu deposited.
Mass = number of moles * molar mass
= 0.158 * 63.5
= 10.035 g of Cu deposited.
Answer:
The amount of Cu²⁺ plated out is 5.02 g.
Explanation:
To solve the problem we must first consider the following equation
Cu²⁺ + 2e -------> Cu (s) ---------- (1)
This equation shows that two moles of electrons are used for the conversion of Cu²⁺. So let us first determine the number of moles of electrons, which can be calculated from the following formula
[tex]n=\frac{I\times t}{F}[/tex]
Here, I is the current in ampere, t is time in seconds and F is the Faraday's constant. Placing the data we get,
[tex]n=\frac{3.53\times 1.2\times 3600}{9.648\times10^{4}}[/tex]
[tex]n=0.158 mol[/tex]
Equation 1 shows that 1 mole of Cu²⁺ consumes two moles of electrons. So, the moles of Cu²⁺ will be half of the moles of electrons.
[tex]n_{Cu^{2+}}=\frac{0.158}{2} mol\\n_{Cu^{2+}}=0.079 mol[/tex]
the mass of Cu²⁺ will be
[tex]g_{Cu^{2+}} = 0.079 mol \times at.mass\\g_{Cu^{2+}} = 0.079 mol \times 63.546 (g/mol)\\g_{Cu^{2+}} = 5.02 g[/tex]
