Kmno4+h2so4+feso4=k2so4+mnso4+fe2(so4)3+h2o balance by partial equation method
Go to Google and pull up Chemteam:main menu and click on oxidation'-reduction. That site gives very good instructions how to do this. Post your work if you need further assistance and tell us exactly what you are having trouble doing.
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To balance the given equation using the partial equation method, we need to follow these steps:
Step 1: Write the unbalanced equation.
K2Cr2O7 + H2SO4 + FeSO4 = K2SO4 + Cr2(SO4)3 + Fe2(SO4)3 + H2O
Step 2: Identify the elements that are changing oxidation state and determine their oxidation numbers on both sides of the equation.
In this case, the elements that change oxidation state are Cr and Fe.
On the reactant side:
Cr: +6 (in K2Cr2O7)
Fe: +2 (in FeSO4)
On the product side:
Cr: +3 (in Cr2(SO4)3)
Fe: +3 (in Fe2(SO4)3)
Step 3: Determine the number of electrons gained or lost by each element.
Cr undergoes a reduction from +6 to +3, which means it gains 3 electrons.
Fe does not change its oxidation state, so it does not gain or lose electrons.
Step 4: Balance the half-reactions for the elements changing oxidation state.
Half-reaction for Cr:
Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O
Half-reaction for Fe:
No change in oxidation state, so no half-reaction is needed.
Step 5: Balance the hydrogen atoms by adding H+ ions to the side where they are insufficient.
On the reactant side, there are 2 H atoms from H2SO4, but on the product side, there are 14 H atoms from H2O and 7 H atoms from 7H2O, totaling 21 H atoms.
Add 10H+ ions to the product side:
Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O + 10H+
Step 6: Balance the oxygen atoms by adding water (H2O) molecules to the side where they are insufficient.
On the product side, there are 7 O atoms from 7H2O, but there are no O atoms on the reactant side. Therefore, we need to add 7 H2O molecules to the reactant side.
Add 7H2O to the reactant side:
K2Cr2O7 + 7H2O + 14H+ + 6e- → 2Cr^3+ + 7H2O + 10H+
Step 7: Balance the charges on both sides by adding electrons (e-) to the side where charge is not balanced.
On the reactant side, we have 6 electrons (from Cr2O7^2-), and on the product side, we have 10 electrons (from 10H+).
Multiply the reactant side by 10 and the product side by 6 to equalize the number of electrons:
10K2Cr2O7 + 70H2O + 140H+ + 60e- → 20Cr^3+ + 70H2O + 100H+
Step 8: Cancel out the water (H2O) molecules that appear on both sides of the equation.
10K2Cr2O7 + 70H2O + 140H+ + 60e- → 20Cr^3+ + 70H2O + 100H+
The water molecules cancel out:
10K2Cr2O7 + 70H2O + 140H+ + 60e- → 20Cr^3+ + 100H+
Step 9: Simplify the equation by canceling out common terms.
Divide the entire equation by a common factor, which in this case is 10:
K2Cr2O7 + 7H2O + 14H+ + 6e- → 2Cr^3+ + 10H+
The balanced equation using the partial equation method is:
K2Cr2O7 + 7H2SO4 + 6FeSO4 → K2SO4 + 2Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O
To balance the given chemical equation using the partial equation method, we need to follow a systematic approach. Here's a step-by-step guide:
Step 1: Split the reaction into half-reactions.
K2MnO4 → K2SO4 (half-reaction 1)
FeSO4 → MnSO4 + Fe2(SO4)3 + H2O (half-reaction 2)
Step 2: Balance the atoms (other than hydrogen and oxygen) in each half-reaction.
For half-reaction 1:
K2MnO4 → K2SO4 (balanced)
For half-reaction 2:
FeSO4 → MnSO4 + 2Fe2(SO4)3 + H2O (unbalanced)
Step 3: Balance the oxygen atoms by adding water molecules to the side that needs them.
For half-reaction 1: No oxygen atoms other than in the initial compound, so it's already balanced.
For half-reaction 2:
FeSO4 → MnSO4 + 2Fe2(SO4)3 + 7H2O (balanced)
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+).
For half-reaction 1: No hydrogen atoms other than in the initial compound.
For half-reaction 2:
FeSO4 + 7H2O → MnSO4 + 2Fe2(SO4)3 + 7H2O (balanced)
Step 5: Balance the charges by adding the appropriate number of electrons (e-).
For half-reaction 1: No charge imbalance.
For half-reaction 2:
FeSO4 + 7H2O → MnSO4 + 2Fe2(SO4)3 + 7H2O + 10e- (balanced)
Step 6: Multiply the half-reactions by appropriate factors to equalize the number of electrons.
Multiply half-reaction 1 by 10:
10K2MnO4 → 10K2SO4
Multiply half-reaction 2 by 1:
10FeSO4 + 70H2O → 10MnSO4 + 20Fe2(SO4)3 + 70H2O + 100e-
Step 7: Now, add the balanced half-reactions together.
Add both half-reactions:
10K2MnO4 + 10FeSO4 + 70H2O → 10K2SO4 + 10MnSO4 + 20Fe2(SO4)3 + 70H2O + 100e-
Step 8: Cancel out the identical species on both sides of the equation.
Final balanced equation:
10K2MnO4 + 10FeSO4 + 70H2O → 10K2SO4 + 10MnSO4 + 20Fe2(SO4)3