If 0.50 L of a 0.60 M SnSO_4 solution is el | vedclass

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(D) Initial moles of $Sn^{2+} = 	ext{Molarity} 	imes 	ext{Volume} = 0.60 \ M 	imes 0.50 \ L = 0.30 \ mol$. The reaction at the cathode is $Sn^{2+}

Vedclass · Accepted Answer

$0.257$

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(D) Initial moles of $Sn^{2+} = 	ext{Molarity} 	imes 	ext{Volume} = 0.60 \ M 	imes 0.50 \ L = 0.30 \ mol$. The reaction at the cathode is $Sn^{2+} + 2e^- ightarrow Sn(s)$. Total charge passed $Q = I 	imes t = 4.60 \ A 	imes (30.0 	imes 60 \ s) = 8280 \ C$. Moles of electrons passed $= \frac{8280}{96500} \approx 0.0858 \ mol$. Moles of $Sn^{2+}$ deposited $= \frac{0.0858}{2} = 0.0429 \ mol$. Remaining moles of $Sn^{2+} = 0.30 - 0.0429 = 0.2571 \ mol$. Final concentration of $Sn^{2+} = \frac{0.2571 \ mol}{0.50 \ L} = 0.514 \ M$. Note: The provided options suggest a calculation error in the original question's source. Based on the standard calculation,the result is $0.514 \ M$.

If $0.50 \ L$ of a $0.60 \ M$ $SnSO_4$ solution is electrolyzed for a period of $30.0 \ min$ using a current of $4.60 \ A$ with inert electrodes,what is the final concentration of $Sn^{2+}$ remaining in the solution? (Atomic weight of $Sn = 119$)

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