Question

In an electroplating cell,0.485g of a metal M are plated out from an acidic solution of MO3 in exactly on hour using current of 1.50A ,What is the molar mass of M ?,What id reduction equation?,what is charge passed? ,how many moles of electrons sre passed?, How many moles of M are deposited?

Answer 1

$Q= it$

Q$=1.5×60×60= 5400c$

If $0.485g$ is electroplated by $5400c,$ then $96500c$ can electroplated;

$(96500/5400)×0.485$

Molar mass$=8.67$g/mol

reduction equation

M3⁺ + 3e^_ $\to$ M_(s)

charge passed$=$ $5400c$

 moles of electrons passed

$=$ 3 moles of electrons

moles of M deposited

Mass$=$ moles $×$ RFM

moles$=$ mass$/$ RFM

$=$ 0.485g$/$ 8.67g/mol

$=$ 0.05594 moles