Question

Complete and balance these half-equations

IO3−→I2 (acidic solution)

HNO3→N2O(g) (acidic solution)

C2O42−→CO2(acidic solution)

Cr2O72−→Cr3+(acidicsolution)

Express your answer as a net ionic equation. Identify all of the phases in your answer

Answer 1

1) Balance the two half-reactions:

2 IO3^- --------> I2 ( oxidation states )
2 (+5) (0)

2IO3^- + 10 e- --------> I2 ( oxidation state get balanced by adding 12 e- to LHS )

2IO3^- + 10e- +12H+ --------> I2 ( negative charge get balanced by adding 12 H+ to LHS )

2IO3^- + 10e- +12 H+ --------> I2 + 6 H2O ( add 6 water molecule to RHS to balance H+ and O )

2) 2HNO3 -------> N2O(g) (acidic solution)

2 (+5 ) 2 (+1)

2HNO3 + 8 e- -------> N2O(g) ( we add 8 e- to the LHS to balance the oxidation state )

  2HNO3 + 8 e- + 8 H+ -------> N2O(g)   ( we add 8 H+ to the LHS to balance the negative charge )

2HNO3 + 8 e- + 8 H+ -------> N2O(g) + 5 H2O   ( add 5 water molecule to RHS to balance H+ and O )

3)   C2O4²⁻ -------> 2 CO2 (acidic solution)

2(+3) 2(+4)

C2O4²⁻   -------> 2 CO2   +2e- ( we add 2 e- to the RHS to get oxidation state get balanced on both side )

C2O4²⁻   -------> 2 CO2 +2e-    ( we donot add H+because equation is already get balanced )

4) Cr2O7²⁻ ----> 2 Cr3+ (acidicsolution)

2( +6) 2 ( +3)

Cr2O7²⁻ + 6 e-   ----> 2 Cr3+     ( oxidation state on both side get balanced by adding 6 e- to LHS )

Cr2O7²⁻ + 6 e- + 14 H+ ----> 2 Cr3+   ( charge on both side get balanced by adding 14 H+ to LHS )

Cr2O7²⁻ + 6 e- + 14 H+ ----> 2 Cr3+ + 7 H2O    ( add 7 water molecule to RHS to balance H+ and O )