posted by Anonymous .
Balance the following equation for a half reaction that occurs in acidic solution. Use e– as the symbol for an electron.
Mo^3+ -> MoO2^2+
Is the answer 2O+2Mo^3+ -> MOO2^2++3e^-?
No. O should be written O2. Mo doesn't balance. Charge on left and right don't balance. The following are the steps. I think I gave you a site last night to show you how to do this. There are easier ways to do this but I prefer the following steps.
Mo^3+ ==> MoO2^2+
1. Determine the oxidation state of the elements and add electrons to the appropriate side to balance the change Mo is changing; on the left is +3. On the right is +6, so we add 3e to the right.
Mo^3+ ==> MoO2^2+ + 3e
2. Count up the charges on the two sides and add H+ (in acid solution or OH in a basic solution) to balance the charge.
+3 on the left; -1 on the right. Add H^+ to the right.
Mo^3+ ==> MoO2^2+ + 3e + 4H^+
3. Ad H2O to the other side to balance H.
Mo^3+ + 2H2O ==> MoO2^2+ + 3e + 4H^+
4. Check everything.
1 Mo left and right
2 O left and right.
4 H left and right
+3 on the left; +2 +4 -3 = 3+ on right.
c. change in oxidation state:
Mo^3+ ==> Mo^6+ + 3e