Question

How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction. (a) Al3+, 1.234 A (b) Ca2+, 22.2 A (c) Cr5+, 37.45 A (d) Au3+, 3.57 A

Answer 1

Explanation:

(a)     It is given that for $Al^{3+}$ value of current is 1.234 A.

Hence, total charge Q = $\frac{3 mol e^{-} \times 96485 C}{1.00 mol e^{-}}$

                                     = 289455 C

Now, we will calculate the time as follows.

                      t = $\frac{Q}{I}$

                        = $\frac{289455 C}{1.234 C/s}$

                        = 234566.4506 s

or,                    = $\frac{234566.4506 s}{3600 s} \times 1 hr$                    

                       = 65.16 hr

(b)    It is given that for $Ca^2+$, value of current is 22.2 A.

Hence, its total charge Q = $\frac{2 mol e^- \times 96485 C}{1.00 mol e^-}$

                                         = 192970 C

Now, we will calculate the time  as follows.

                            t = $\frac{Q}{I}$

                              = $\frac{192970 C}{22.2 C/s}$

                              = 8692.34 s

or,                          = 2.4 hr

(c)   It is given that for $Cr^5+$, value of current is 37.45 A.

Hence, the total charge Q = $\frac{5 mol e^{-} \times 96485 C}{1.00 mol e^{-}}$

                                        = 482425 C

Now, we will calculate the time  as follows.

                      t = \frac{Q}{I}[/tex]

                        = \frac{482425 C}{37.45 C/s}[/tex]

                       = 12881.8 s

or,                    = 3.6 hr  

(d)    It is given that for $Au^{3+}$, value of current is 3.57 A.

Hence, the total charge Q = $\frac{3 mol e^{-} \times 96485 C}{1.00 mol e^{-}}$

                                          = 289455 C

Now, we will calculate the time as follows.

                      t = $\frac{Q}{I}$

                        = $\frac{289455 C}{3.57 C/s}$

                       = 81079.83 s

                       = 22.5 hr