Question
Balance the following ionic equations. $\text{Cr}_2\text{O}^{2-}_7+\text{H}^++\text{I}^-\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{Cr}^{3+}+\text{I}_2+\text{H}_2\text{O}$
✓
Answer
Dividing the equation into two half reactions:
Oxidation half reaction: $\text{I}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{I}_2$
Reduction half reaction: $\text{Cr}_2\text{O}^{2-}_7\xrightarrow{ \ \ \ \ \ \ \ }\text{Cr}^{3+}$
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
$\text{I}^{-}\xrightarrow{ \ \ \ \ \ \ \ }\text{I}_2$
$2\text{I}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{I}_2$
$\text{2I}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{I}_2+2\text{e }^-...(\text{i})$
Reduction half reaction
$\text{Cr}_2\text{O}^{2-}_7\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Cr}^{3+}$
$\text{Cr}_2\text{O}^{2-}_7\xrightarrow{ \ \ \ \ \ \ \ }2\text{Cr}^{3+}$
$\text{Cr}_2\text{O}^{2-}_7+6\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{Cr}^{3+}$
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{Cr}^{3+}+7\text{H}_2\text{O }...(\text{ii})$
To balance the electrons, multiply eq. (i) by 3 and add to eq. (ii)
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{I}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{Cr}^{3+}+3\text{I}_2+7\text{H}_2\text{O}$
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\begin{array}{l}
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\end{array}
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\begin{array}{l}
2 C(s)+2 H_2(g)=C_2 H_4(g) ; \Delta H=52.6 kJ \\
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\end{array}
$
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|
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