How do I balance a redox reaction for NaI + H2SO4 = H2S + I2 + Na2SO4 + H2O? I don't understand how to balance the electron loss for Iodine.
To balance a redox reaction, you need to assign oxidation states to each element in the equation and then balance the number of electrons gained and lost. Let's go through the process step by step:
Step 1: Assign oxidation states to each element:
In NaI, sodium (Na) has an oxidation state of +1, and iodine (I) has an oxidation state of -1.
In H2SO4, hydrogen (H) has an oxidation state of +1, oxygen (O) has an oxidation state of -2, and sulfur (S) has an oxidation state of +6.
In H2S, hydrogen (H) has an oxidation state of +1, and sulfur (S) has an oxidation state of -2.
In I2, iodine (I) has an oxidation state of 0.
In Na2SO4, sodium (Na) has an oxidation state of +1, oxygen (O) has an oxidation state of -2, and sulfur (S) has an oxidation state of +6.
In H2O, hydrogen (H) has an oxidation state of +1, and oxygen (O) has an oxidation state of -2.
Step 2: Identify the atoms undergoing oxidation and reduction:
In this reaction, iodine (I) is being oxidized from -1 to 0, and sulfur (S) is being reduced from +6 to -2.
Step 3: Balance the atoms, excluding oxygen and hydrogen:
Let's first balance the non-hydrogen and non-oxygen atoms using coefficients:
NaI + H2SO4 = H2S + I2 + Na2SO4
Step 4: Balance oxygen atoms:
To balance the number of oxygen atoms, add H2O on the appropriate side of the equation:
NaI + H2SO4 = H2S + I2 + Na2SO4 + H2O
Step 5: Balance hydrogen atoms:
To balance the number of hydrogen atoms, add H+ ions (protons) on the appropriate side of the equation:
NaI + H2SO4 + 2H+ = H2S + I2 + Na2SO4 + H2O
Step 6: Balance the charges:
To balance the overall charge, add electrons (e-) on the appropriate side of the equation:
NaI + H2SO4 + 2H+ = H2S + I2 + Na2SO4 + H2O + 2e-
Step 7: Balance the electron transfer:
To balance the number of electrons gained and lost, you need to make the total number of electrons transferred equal. In this case, 2 electrons are being lost during oxidation (from iodine), so we need to add 2 electrons on the left side:
2NaI + H2SO4 + 2H+ = H2S + I2 + Na2SO4 + H2O + 2e-
The final balanced redox equation for NaI + H2SO4 = H2S + I2 + Na2SO4 + H2O is:
2NaI + H2SO4 + 2H+ = H2S + I2 + Na2SO4 + H2O + 2e-
First we note that we have 1 I on the left and 2 on the right. In order to compare apples with apples we need to have the same number; therefore, we stick a 2 on the left so we start with the same number of I atoms.
2I^- ==>I2
So we have changed from -2 (for two I atoms) to zero (for two I atoms). That is a loss of two electrons.
2I^- ==> I2 + 2e
That first step is almost never addressed in redox instructions and it is the difference between success and failure. It is common to do this with I^- ==> I2, Cr2O7^2- ==> Cr^3+, Cl2 ==> Cl^-, S2O4^2- ==> S4O6^2-, C2O4^2- ==> CO2, etc.