To determine the quantity of electrical charge required to plate 1.386 moles of chromium (Cr) from an acidic solution of potassium dichromate (K2Cr2O7), we first need to establish the balanced half-reaction for the reduction of chromium ions. This will help us understand how many electrons are involved in the process.
Balanced Half-Reaction for Chromium Reduction
The reduction of chromium from K2Cr2O7 in an acidic medium can be represented by the following half-reaction:
Half-Reaction
Cr2O72- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O
In this half-reaction, one mole of dichromate ion (Cr2O72-) is reduced to produce two moles of chromium ions (Cr3+). Notably, this process requires the transfer of 6 electrons (e-) for every 2 moles of chromium produced.
Calculating the Charge Required
Next, we need to calculate the total charge needed to reduce 1.386 moles of chromium. Since 2 moles of Cr require 6 moles of electrons, we can set up a proportion to find out how many moles of electrons are needed for 1.386 moles of Cr:
- For 2 moles of Cr: 6 moles of e-
- For 1.386 moles of Cr: x moles of e-
Using the ratio:
x = (1.386 moles Cr) × (6 moles e- / 2 moles Cr) = 4.158 moles e-
Finding the Total Charge
Now that we know we need 4.158 moles of electrons, we can calculate the total charge (Q) using Faraday's constant (F), which is approximately 96,485 C/mol. The formula to find the charge is:
Q = n × F
Where:
- Q = total charge in coulombs (C)
- n = number of moles of electrons
- F = Faraday's constant (96,485 C/mol)
Substituting the values we have:
Q = 4.158 moles e- × 96,485 C/mol = 401,200.93 C
Final Result
Thus, the quantity of electrical charge needed to plate 1.386 moles of chromium from an acidic solution of K2Cr2O7 is approximately 401,201 coulombs. This calculation illustrates the relationship between moles of substance, the number of electrons involved in the reduction process, and the total charge required for electroplating.