Return to balancing half-reactions in acid 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