Question

Amount of $Cu$ deposited from $Cu{SO}_4\left(aq\right)$ when $5amp$ current is passed for $965$ seconds through it, is ($Cu=63.5g{mol}^{-1}$)

• A. $1.6g$
• B. $3.2g$
• C. $1.45g$
• D. $2.9g$

Answer 1

Answer: (A)

$1.6g$

5 amperes is 5 coulombs per second, 5C/s

So the total charge in 965 seconds is $Q=5C/s\times 965s=4825C$

Then the number of moles of copper plated out (n) is:

$n=\frac{Q}{zF}$

where z is the number of electrons in the half-cell reaction and F is the Faraday constant = 96,485/mol

So, $n=\frac{4825}{(2\times 96485)}=0.025mol$

And this is $63.546\times 0.025=1.58g$

So 1.58grams are plated deposited at the cathode.