### Practice problem answersAcid solution

1) Re ---> ReO2

Step One: Balance the atom being reduced/oxidized: Re ---> ReO2
Step Two: Balance the oxygens: 2H2O + Re ---> ReO2
Step Three: Balance the hydrogens: 2H2O + Re ---> ReO2 + 4H+
Step Four: Balance the total charge: 2H2O + Re ---> ReO2 + 4H+ + 4e¯

2) Cl2 ---> HClO

1) Cl2 ---> 2HClO (chlorine is balanced)
2) 2H2O + Cl2 ---> 2HClO (now there are 2 oxygens on each side)
3) 2H2O + Cl2 ---> 2HClO + 2H+ (2H in HClO plus 2H+ makes 4 hydrogens)
4) 2H2O + Cl2 ---> 2HClO + 2H+ + 2e¯ (zero charge on the left; +2 from the hydrogen ions so 2 electrons gives the -2 charge required to make zero on the right)

3) NO3¯ ---> HNO2

2e¯ + 3H+ + NO3¯ ---> HNO2 + H2O

4) H2GeO3 ---> Ge

4e¯ + 4H+ + H2GeO3 ---> Ge + 3H2O

5) H2SeO3 ---> SeO42¯

H2O + H2SeO3 ---> SeO42¯ + 4H+ + 2e¯

6) Au ---> Au(OH)3

3H2O + Au ---> Au(OH)3 + 3H+ + 3e¯

7) H3AsO4 ---> AsH3

8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O

8) H2MoO4 ---> Mo

6e¯ + 6H+ + H2MoO4 ---> Mo + 4H2O

9) NO ---> NO3¯

2H2O + NO ---> NO3¯ + 4H+ + 3e¯

10) H2O2 ---> H2O

2e¯ + 2H+ + H2O2 ---> 2 H2O