Question

Balance the following ionic equations:
(A) $ C{r_2}O_7^{2 - } + {H^ + } + {I^ - } \to C{r^{3 + }} + {I_2} + {H_2}O $
(B) $ C{r_2}O_7^{2 - } + F{e^{2 + }} + {H^ + } \to C{r^{3 + }} + F{e^{3 + }} + {H_2}O $
(C) $ MnO_4^ - + SO_3^{2 - } + {H^ + } \to M{n^{2 + }} + SO_4^{2 - } + {H_2}O $
(D) $ MnO_4^ - + B{r^ - } + {H^ + } \to M{n^{2 + }} + B{r_2} + {H_2}O $

Answer 1

Hint: All the reactions in the given question are redox reactions. To answer this question, you must first determine the reduction half- cell and oxidation half-cell reactions before proceeding to balance the equations.

Complete step by step solution
(i) Dichromate ion is a very good oxidizing agent with chromium present in the oxidation state.
We are given the reaction, $ C{r_2}O_7^{2 - } + {H^ + } + {I^ - } \to C{r^{3 + }} + {I_2} + {H_2}O $
First we balance all the atoms other than hydrogen and oxygen in the reaction. We get,
 $ C{r_2}O_7^{2 - } + {H^ + } + 2{I^ - } \to 2C{r^{3 + }} + {I_2} + {H_2}O $
Next, we write both the half- cell reactions.
Reduction half- cell: $ C{r_2}O_7^{2 - } \to 2C{r^{3 + }} $ : each chromium atom gains three electrons. Overall 6 electrons are lost.
Oxidation half cell: $ 2{I^ - } \to {I_2} $ : Each iodide ion loses an electron. Overall two electrons are gained.
Multiplying the oxidation reaction by three and adding, we get,
 $ C{r_2}O_7^{2 - } + 6{I^ - } \to 2C{r^{3 + }} + 3{I_2} $
Balancing the charges, and hydrogen and oxygen atoms.
 $ C{r_2}O_7^{2 - } + 14{H^ + } + 6{I^ - } \to 2C{r^{3 + }} + 3{I_2} + 7{H_2}O $
(ii) We are given the reaction, $ C{r_2}O_7^{2 - } + F{e^{2 + }} + {H^ + } \to C{r^{3 + }} + F{e^{3 + }} + {H_2}O $
First we balance all the atoms other than hydrogen and oxygen in the reaction. We get,
 $ C{r_2}O_7^{2 - } + F{e^{2 + }} + {H^ + } \to 2C{r^{3 + }} + F{e^{3 + }} + {H_2}O $
Next, we write both the half- cell reactions.
Reduction half- cell: $ C{r_2}O_7^{2 - } \to 2C{r^{3 + }} $ : each chromium atom gains three electrons. Overall 6 electrons are lost.
Oxidation half cell: $ F{e^{2 + }} \to F{e^{3 + }} $ : Each ferrous ion loses an electron.
Multiplying the oxidation reaction by six and adding, we get,
 $ C{r_2}O_7^{2 - } + 6F{e^{2 + }} \to 2C{r^{3 + }} + 6F{e^{3 + }} $
Balancing the charges, and hydrogen and oxygen atoms.
 $ C{r_2}O_7^{2 - } + 6F{e^{2 + }} + 14{H^ + } \to 2C{r^{3 + }} + 6F{e^{3 + }} + 7{H_2}O $
(iii) We are given the reaction, $ MnO_4^ - + SO_3^{2 - } + {H^ + } \to M{n^{2 + }} + SO_4^{2 - } + {H_2}O $
Writing the half- cell reactions as,
 $ MnO_4^ - \to M{n^{2 + }} $ : Manganese gains 5 electrons
 $ 2B{r^ - } \to B{r_2} $ : Sulphur loses 2 electrons
Cross multiplying the transferred electrons and adding, we get,
 $ 2MnO_4^ - + 5SO_3^{2 - } \to 2M{n^{2 + }} + 5SO_4^{2 - } $
Balancing the charges, and hydrogen and oxygen atoms.
 $ 2MnO_4^ - + 5SO_3^{2 - } + 6{H^ + } \to 2M{n^{2 + }} + 5SO_4^{2 - } + 3{H_2}O $
(iv) We are given the reaction, $ MnO_4^ - + B{r^ - } + {H^ + } \to M{n^{2 + }} + B{r_2} + {H_2}O $
Writing the half- cell reactions as,
 $ MnO_4^ - \to M{n^{2 + }} $ : Manganese gains 5 electrons
 $ 2B{r^ - } \to B{r_2} $ : 2 electrons are lost
Cross multiplying the transferred electrons and adding, we get,
 $ 2MnO_4^ - + 10B{r^ - } \to 2M{n^{2 + }} + 5B{r_2} $
Balancing the charges, and hydrogen and oxygen atoms.
 $ 2MnO_4^ - + 10B{r^ - } + 16{H^ + } \to 2M{n^{2 + }} + 5B{r_2} + 8{H_2}O $

Note
A reaction in which one species is oxidized while the other is reduced is known as a redox (reduction- oxidation reaction). Balancing a normal chemical equation is different than balancing a redox reaction.