2MnO4 + H + 5HSO3 = 2Mn + 5HSO4 + 3H2O
I was asked to write the half equations for both the oxidized and reduced reactions.
Oxidized
• 2Mn2+ + 8H2O ---> 2MnO4- + 16H+ +10e-
Reduced
• 5SO42- + 15H+ +10e- -----> 5HSO3- + 5H2O
Are these half equations correct?
They are oxidzed and reduced equations that you wrote but you didn't write the right ones. In the equation it's MnO4^- that is reduced to Mn^2+ and the SO3^2- is oxidized to SO4^2-
You wrote the oxidized and reduced equations for the REVERSE equation.
Oxidized
• 5SO42- + 15H+ +10e- -----> 5HSO3- + 5H2O
Reduced
• 2Mn2+ + 8H2O ---> 2MnO4- + 16H+ +10e-
I am correct now?
The half equations you provided are not correct. Let's go through the process of balancing the half equations for the oxidation and reduction reactions.
Oxidation Half Equation:
In this reaction, Mn is being oxidized from +2 to +7.
To balance the half equation, we need equal numbers of Mn on both sides. Since there are 2 Mn atoms on the left side, we will balance it by multiplying MnO4- by 2.
2MnO4- + __H2O ---> 2MnO4- + __H+ + __e-
Next, let's balance the oxygens by adding H2O to the other side.
2MnO4- + __H2O ---> 2MnO4- + __H+ + __e- + __H2O
Now let's balance the hydrogens by adding H+ ions.
2MnO4- + __H2O ---> 2MnO4- + __H+ + __e- + __H2O
Finally, let's balance the charges by adding electrons (e-).
2MnO4- + __H2O ---> 2MnO4- + 8H+ + 10e- + __H2O
Thus, the balanced oxidation half equation is:
2Mn2+ + 8H2O ---> 2MnO4- + 8H+ + 10e-
Reduction Half Equation:
In this reaction, SO3 is being reduced from +6 to +4.
To balance the half equation, we need equal numbers of SO3 on both sides. Since there are 5 SO3 on the left side, we will balance it by multiplying HSO3- by 5.
__SO3 + __H2O ---> __HSO3- + __H+ + __e-
Next, let's balance the oxygens by adding H2O to the other side.
__SO3 + __H2O ---> __HSO3- + __H+ + __e- + __H2O
Now let's balance the hydrogens by adding H+ ions.
__SO3 + __H2O ---> __HSO3- + 5H+ + __e- + __H2O
Finally, let's balance the charges by adding electrons (e-).
__SO3 + __H2O ---> __HSO3- + 5H+ + 10e- + __H2O
Thus, the balanced reduction half equation is:
__SO3 + __H2O ---> __HSO3- + 5H+ + 10e-
Please note that there may be some additional balancing needed in terms of the coefficients of the species involved, depending on the context of your question.
Yes, both half equations you provided are correct for the reaction 2MnO4 + H + 5HSO3 = 2Mn + 5HSO4 + 3H2O.
To determine the half equations, you need to identify the elements or compounds that are being oxidized and reduced. In this reaction, manganese (Mn) is being reduced from +7 oxidation state in MnO4- to +2 oxidation state in Mn2+. Therefore, the half equation for the reduction of Mn is:
2Mn2+ + 8H2O ---> 2MnO4- + 16H+ + 10e-
On the other hand, sulfur (S) is being oxidized from +4 oxidation state in HSO3- to +6 oxidation state in HSO4-. Therefore, the half equation for the oxidation of S is:
5SO42- + 15H+ +10e- -----> 5HSO3- + 5H2O
It's important to note that the number of electrons transferred in the oxidation and reduction half equations must be balanced in order to maintain overall charge neutrality. In this case, the numbers of electrons on both sides of the half equations are balanced.