Question
Balance the following ionic equations. $\text{Cr}_2\text{O}^{2-}_7+\text{Fe}^{2+}+\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Cr}^{3+}+\text{Fe}^{3+}+\text{H}_2\text{O}$
✓
Answer
Step-1: Separate the equation into two half reactions.
The oxidation number of various atoms are shown below:
$\stackrel{+6 \ \ -2 \ \ \ \ \ \ \ }{\text{Cr}_2\text{O}^{2-}_7}+\stackrel{+2 \ \ \ \ \ \ \ }{\text{Fe}^{2+}}+\stackrel{+1 \ \ \ }{\text{H}^+}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+3}{\text{Cr}^{3+}}+\stackrel{+3}{\text{Fe}^{3+}}+\stackrel{+1 \ \ -2}{\text{H}_2\text{O}}$
In this case, chromium undergose reduction, oxidation number decreases from $+6(\text{in }\text{Cr}_2\text{O}^{2-}_7)\text{ to}+3(\text{in Cr}^{3+})$
$Fe^{2+} (O.N. = +2)$ changes to $Fe^{3+}$ (O.N. = +3). The species undergoing oxidation and reduction are:
Oxidation: $\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }\text{Fe}^{3+}$
reduction: $\text{Cr}_2\text{O}^{2-}_7\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Cr}^{3+}$
Step-2: Balance each half reaction separately as:
- $\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Fe}^{3+}$
- Balance all atoms other than H and O. This step is not needed, because, it is already balanced.
- The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as: $\text{Fe}^{2+}\xrightarrow{\ \ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-$
- Charge is already balanced.
- No need to add H or O.
$\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-...(\text{i})$
Consider the second half equation
$\text{Cr}_2\text{O}^{2-}_7\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Cr}^{3+}$
- Balance the atoms other than H and O.
- The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three electrons. Since there are two Cr atoms, add $6e^–$ on the left.
- Since the reaction takes place in acidic medium add $14H^+$ on the left to equate the net charge on both sides.
- To balance FI atoms, add $7H_2O$ molecules on the right.
This is the balanced half equation.
Step-3: Now add up the two half equations. Multiply eq. (i) by 6 so that electrons are balanced.
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-\times6\\ \ \ \ \ \ \ \ \ \text{Cr}_2\text{O}^{2-}_7+6\text{e}^-+14\text{H}\xrightarrow{ \ \ \ \ \ \ \ \ \ }2\text{Cr}^{3+}+7\text{H}_2\text{O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \underline{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\6\text{Fe}^{2+}+\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }6\text{Fe}^{3+}+2\text{Cr}^{3+}+7\text{H}_2\text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
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