Question 21.11: Applying the Relationship Among Current, Time, and Amount of......

Plan To find the current, we divide charge by time, so we need to find the charge. We write the half-reaction for the Cr^{3+} reduction to get the amount (mol) of e^− transferred per mole of Cr. To find the charge, we convert the mass of Cr needed (0.86 g) to amount (mol) of Cr. Then, we use the Faraday constant (9.65×10^4 \text{ C/mol }e^−) to find the charge and divide by the time (12.5 min, converted to seconds) to obtain the current (see the road map (Fig 21.11)).

Solution Writing the balanced half-reaction [Cr_2(SO_4)_3 contains the Cr^{3+} ion]:
               Cr^{3+}(aq)  +  3e^−  ⟶  Cr(s)
Combining steps to find amount (mol) of e^− transferred for the mass of Cr needed:

Calculating the charge using the Faraday constant as a conversion factor between moles of electrons transferred and charge:

Calculating the current:
             \text{Current (A) }=  \frac{\text{charge (C)}}{\text{time (s)}}  =  \frac{4.8×10^3  C}{12.5 \min}  ×  \frac{1  \min}{60  s}  =  6.4  C/s  =  6.4  A
Check Rounding gives

Then
            (5×10^{−2} \text{ mol }e^−)(∼1×10^5 \text{ C/mol }e^−)  =  5×10^3  C
and

Comment In order to introduce Faraday’s law, we have neglected some details about actual electroplating. In practice, electroplating chromium has an efficiency of only 30–40% and must be run at a specific temperature range for the plate to appear bright. Nearly 10,000 metric tons (2×10^8 mol) of chromium are used annually for electroplating.