Question

the dichromate ion, $mathrm{cr_2o_7^{2-}}$, is reduced by the chloride ion in acidic solution. balance the dichromate half-reaction from the skeleton equation.
$mathrm{cr_2o_7^{2-} \
ightarrow cr^{3+}}$
how many electrons are required to balance this half-reaction?

Explanation:

Step1: Balance Cr atoms

In the skeleton equation $\ce{Cr2O7^{2-} -> Cr^{3+}}$, there are 2 Cr atoms on the left and 1 on the right. So we multiply $\ce{Cr^{3+}}$ by 2:
$\ce{Cr2O7^{2-} -> 2Cr^{3+}}$

Step2: Balance O atoms with $\ce{H2O}$

There are 7 O atoms on the left. To balance O, we add 7 $\ce{H2O}$ on the right:
$\ce{Cr2O7^{2-} -> 2Cr^{3+} + 7H2O}$

Step3: Balance H atoms with $\ce{H+}$

There are 14 H atoms in 7 $\ce{H2O}$ on the right. So we add 14 $\ce{H+}$ on the left:
$\ce{Cr2O7^{2-} + 14H+ -> 2Cr^{3+} + 7H2O}$

Step4: Balance charge with electrons

Calculate the charge on each side. Left side: charge of $\ce{Cr2O7^{2-}}$ is -2, charge of 14 $\ce{H+}$ is +14, total is -2 + 14 = +12. Right side: charge of 2 $\ce{Cr^{3+}}$ is 2*3 = +6, charge of 7 $\ce{H2O}$ is 0, total is +6. The difference in charge is +12 - +6 = +6. To balance the charge, we need to add 6 electrons (each electron has -1 charge) on the left (since left side is more positive):
$\ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$

Answer:

6