how many hours would it take to deposit 12.40 g of cr from a solution of crcl3 using a current of 10.00 amp?

Explanation:

Step1: Determine the molar - mass of Cr

The molar - mass of Cr (chromium) is $M = 52.00\ g/mol$.

Step2: Calculate the number of moles of Cr deposited

The number of moles $n$ of Cr is calculated using the formula $n=\frac{m}{M}$, where $m = 12.40\ g$ and $M = 52.00\ g/mol$. So, $n=\frac{12.40\ g}{52.00\ g/mol}=0.2385\ mol$.

Step3: Determine the half - reaction and the number of electrons transferred

The half - reaction for the deposition of Cr from $CrCl_3$ is $Cr^{3 +}+3e^-\rightarrow Cr$, so $z = 3$ (the number of electrons transferred per atom of Cr).

Step4: Calculate the total charge required

The total charge $Q$ required is given by $Q=n\times z\times F$, where $F = 96485\ C/mol$ (Faraday's constant). So, $Q=0.2385\ mol\times3\times96485\ C/mol = 69179.775\ C$.

Step5: Calculate the time required

We know that $I=\frac{Q}{t}$, where $I = 10.00\ A$ (current). Rearranging for $t$, we get $t=\frac{Q}{I}$. So, $t=\frac{69179.775\ C}{10.00\ A}=6917.9775\ s$.

Step6: Convert time from seconds to hours

Since $1\ h = 3600\ s$, then $t=\frac{6917.9775\ s}{3600\ s/h}=1.92\ h$.

Answer:

$1.92$ h