Use half reaction to balance:
CuS + NO3^- = Cu(NO3)2 + NO2 + SO2
S^2- = SO2 S Oxidized
Cu^2+ + NO3^- = NO2 + Cu(NO3)2 N Red
After balancing the atoms & balancing the charges, I cannot get all the correct coefficients
To balance the given chemical equation using half-reactions, we need to balance the charge and the number of atoms of each element on both sides of the equation.
Let's start with the oxidation half-reaction (where sulfur S is oxidized):
CuS → SO2
First, let's balance the sulfur atoms by adding a coefficient of 1/2 in front of the SO2:
CuS → 1/2 SO2
To balance the oxidation state of sulfur, we need to add electrons (e^-) to the right side. The number of electrons needed can be determined by comparing the oxidation states of sulfur in CuS (0) and SO2 (+4).
CuS + 4e^- → 1/2 SO2
Now let's move on to the reduction half-reaction (where nitrogen N is reduced):
NO3^- → NO2
To balance the nitrogen atoms, we need to add a coefficient of 2 in front of the NO2:
NO3^- → 2 NO2
To balance the charge, we need to add electrons (e^-) to the left side. The number of electrons needed can be determined by comparing the charges of NO3^- (-1) and NO2 (0).
1 NO3^- + 4e^- → 2 NO2
Now, we need to balance the number of electrons gained and lost in both half-reactions. To do this, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
2 CuS + 8e^- → SO2
2 NO3^- + 8e^- → 4 NO2
Now, add the two balanced half-reactions together:
2 CuS + 8e^- + 2 NO3^- + 8e^- → SO2 + 4 NO2
Simplify the equation:
2 CuS + 2 NO3^- → SO2 + 4 NO2
Finally, you can see that the balanced equation is:
2 CuS + 2 NO3^- → SO2 + 4 NO2
Note that you can multiply the entire equation by a common factor to get rid of any fractions. However, in this case, the equation is already balanced without fractional coefficients.