Question

qn below

Answer 1

You familiar with
\[I=\frac{q_e}{t} ~~~~ and ~~~~ N_A=\frac{q_e}{F}\]?

Answer 2

What we got in the solution is
\[Cu^{2+}, SO_4^{2-},H_2O\]
Reaction of copper at the cathode will be
\[Cu^{2+}+2e^{-} \rightarrow Cu\]
We'll leave that first
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Then using \(\large I=\frac{q_e}{t}\)
We can find \(\large q_e\), then use this to find \(N_A\) using formula \(\large N_A=\frac{q_e}{F}\)
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Then use stoichiometry to find the mass, remeber that 1 mol of Cu requires 2 mols of electron.

Answer 3

Oh and F=Faraday constant=\(9.65\times 10^{4}Cmol^{-1}e^{-1}\)