A 250.0 mL sample of a 0.20 M Cr^{3+} is e | vedclass

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(D) The number of moles of $Cr^{3+}$ electrolyzed is calculated by the change in concentration:$n = V 	imes \Delta M = 250 	imes 10^{-3} \ L 	imes

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) The number of moles of $Cr^{3+}$ electrolyzed is calculated by the change in concentration:$n = V 	imes \Delta M = 250 	imes 10^{-3} \ L 	imes (0.20 \ M - 0.10 \ M) = 0.025 \ mol$.According to Faraday's law of electrolysis,the amount of substance deposited is given by:$m = \frac{M_{atomic}}{n_{factor} 	imes F} 	imes I 	imes t$Here,$n_{factor} = 3$ for the reduction $Cr^{3+} + 3e^- ightarrow Cr$.Substituting the values:$0.025 \ mol = \frac{1}{3 	imes 96500 \ C/mol} 	imes 96.5 \ A 	imes t$$0.025 = \frac{96.5}{3 	imes 96500} 	imes t$$0.025 = \frac{1}{3000} 	imes t$$t = 0.025 	imes 3000 = 75 \ s$.

$A$ $250.0 \ mL$ sample of a $0.20 \ M \ Cr^{3+}$ is electrolyzed with a current of $96.5 \ A$. If the remaining $[Cr^{3+}]$ is $0.1 \ M$,the duration of the process is $............. \ s$ (Atomic weight of $Cr = 52$).

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