Question

A current of $1.40$ ampere is passed through $500$mL of $0.180$M solution of zinc sulphate for $200$ seconds. The molarity of $Z{n}^{2+}$ ions after deposition of zinc is:

• A. $0.154$M
• B. $0.177$M
• C. $2$M
• D. $0.180$M

Answer 1

Answer: (B)

$0.177$M

$Z{n}^{2+}+2e^{-}⟶Zn$

$Faraday=\frac{Charge}{96500}=\frac{1.40\times 200}{96500}=2.90\times {10}^{-3}$ $mol$

$\therefore Zn$ deposited $=\frac{2.90\times {10}^{-3}}{2}=1.45\times {10}^{-3}$ $mol$

Now, $mol$ of $Z{n}^{2+}$ initially $=0.180\times 0.5$

$Molarity\times V=0.09$ $mol$

$mol$ of $Z{n}^{2+}$ left after deposition $=0.09-0.00145=0.08855$ $mol$

$Molarity=\frac{Moles}{Volume}=\frac{0.08855}{0.5}=0.177$ $M$

Hence option (B) is correct.