Balance the following ionic equations
As step (i), we will be generating the unbalanced skeleton, i.e.,
As step (ii), we will have to Identify the reactants which will get oxidised or reduced and then write the half-cell reaction for them. This will be done along with the electron transfers, as it will carefully make equal numbers of atoms in oxidised and reduced redox couples.
Oxidation is and Reduction is
As step (iii), we will balance the number of atoms in the oxidised and reduced redox couples.
Oxidation is and Reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
The oxidation is and reduction is
As part of step (v), we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction to balancing the oxygen.
Hence, oxidation is and
reduction is
As part of step (vi), we will be making the electron loss and gain equal in both the reduction and oxidation reaction. We will now multiply the oxidation reaction by 3, which will give us:
The oxidation occurs as and
the reduction as
Now, as part of step (vii), we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii), we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
Now, we know that the charge on the LHS is +6, which is same as the RHS of the equation, thereby ensuring that the reaction is balanced.
As step (i), we will be generating the unbalanced skeleton, i.e.,
As step (ii), we will have to Identify the reactants which will get oxidised or reduced and then write the half-cell reaction for them. This will be done along with the electron transfers, as it will carefully make equal numbers of atoms in oxidised and reduced redox couples.
Thus, oxidation is and the reduction is
As step (iii) we will balance the number of atoms in the oxidised and reduced redox couples.
Thus, oxidation is and the reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is
reduction is
As part of step (v), we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction to balancing the oxygen.
Thus, oxidation is
reduction is
As part of step (vi), we will be making the electron loss and gain equal in both the reduction and oxidation reaction.
Thus, oxidation is
reduction is
Now, as part of step (vii), we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii), we will simplify the equation. This will be done by eliminating the similar terms present on both sides of the equation.
We will now verify whether all the charges are balanced.
So, the LHS = and the RHS =
Thus, the sum total is same on both the sides, therefore the solved reaction is right.
As step (i), we will be generating the unbalanced skeleton, i.e.,
Here, Mn undergoes reduction and S undergoes oxidation.
As step (ii), we will have to Identify the reactants which will get oxidised or reduced and then write the half-cell reaction for them. This will be done along with the electron transfers, as it will carefully make equal numbers of atoms in oxidised and reduced redox couples.
Thus, oxidation is and reduction are
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.
Thus, oxidation is and reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is and
reduction is
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Thus, oxidation is and
reduction is
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
Thus, oxidation is and
reduction is
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
We will now verify whether all the charges are balanced.
So, LHS = and the RHS =
Equal charges on both sides imply towards a balanced equation.
As step (i), we will be generating the unbalanced skeleton, i.e.,
As step (ii), we will have to Identify the reactants which will get oxidised or reduced and then write the half-cell reaction for them. This will be done along with the electron transfers, as it will carefully make equal numbers of atoms in oxidised and reduced redox couples.
Thus, oxidation is and reduction is
As step (iii), we will balance the number of atoms in the oxidised and reduced redox couples.
As we can see, the atoms are balanced, so the reaction is going to be the same as the previous step.
Thus, oxidation is and reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is and
reduction is
As part of step (v), we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction to balancing the oxygen.
Thus, the oxidation is and
reduction is
As part of step (vi), we will be making the electron loss and gain equal in both the reduction and oxidation reaction.
Thus, oxidation is and
reduction is
Now, as part of step (vii), we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii), we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
Thus, the charge on LHS and RHS are equal.