Using redox methods, write the balanced equations for each of the reactions in the appropriate medium.
A) Ca(s) + Cr2O7(Charge 2-) (aq) --> Ca2+ (aq) + 2Cr3+ in acid medium
B) Fe(CN)6(charge 4-) (aq) + Ce(charge 4+) --> Ce(OH)3(s) + Fe(OH)3(s) + CO3(charge 2-) (aq) + NO3charge -)
I have a test on this in 5 days and I'm so lost, so if you could please explain the step by step process you could use for any problems like this that would be AMAZING. Thank you in advance!
Here is a good site that shows how to calculate the oxidation states and how to use them to balance a redox equation.
http://www.chemteam.info/Redox/Redox.html
Thank you so much for your help! This is a good site.
Sure! I'll be happy to help you understand the step-by-step process for balancing redox reactions. Redox reactions involve the transfer of electrons between species, resulting in changes in oxidation states.
Let's go through each problem one by one.
A) Ca(s) + Cr2O7(2-) (aq) → Ca2+ (aq) + 2Cr3+ (aq) in acid medium
Step 1: Identify the oxidation states of each element in the reaction.
Ca: The oxidation state of an element in its pure form is 0, so Ca(s) has an oxidation state of 0.
Cr: The oxidation state of Cr in Cr2O7(2-) is +6 since the sum of the oxidation states of all atoms in the compound is -2.
O: The oxidation state of oxygen is -2.
Since the overall charge of Cr2O7(2-) is 2-, the sum of the oxidation states must add up to 2-. So, 2(6) + 7(-2) = -2.
Step 2: Determine which elements are being oxidized and reduced.
In this reaction, Ca is being oxidized from an oxidation state of 0 to +2, and Cr is being reduced from +6 to +3.
Step 3: Write the two half-reactions, one for oxidation and one for reduction.
Oxidation: Ca(s) → Ca2+ (aq) + 2e- (This half-reaction shows the loss of electrons by the Ca atom.)
Reduction: Cr2O7(2-) (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 7H2O (This half-reaction shows the gain of electrons by the Cr atoms.)
Step 4: Equalize the number of electrons in both half-reactions.
Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 2 to equalize the number of electrons.
Oxidation: 6Ca(s) → 6Ca2+ (aq) + 12e-
Reduction: 2Cr2O7(2-) (aq) + 14H+ (aq) + 12e- → 4Cr3+ (aq) + 7H2O
Step 5: Balance the atoms other than O and H.
There are already 6 Ca atoms and 2 Cr atoms on both sides, so they are balanced.
Step 6: Balance the oxygens by adding H2O molecules.
Oxidation: 6Ca(s) → 6Ca2+ (aq) + 12e-
Reduction: 2Cr2O7(2-) (aq) + 14H+ (aq) + 12e- → 4Cr3+ (aq) + 7H2O
To balance oxygen atoms, add 7 H2O to the oxidation half-reaction.
Step 7: Balance the hydrogens by adding H+ ions.
Oxidation: 6Ca(s) → 6Ca2+ (aq) + 12e-
Reduction: 2Cr2O7(2-) (aq) + 14H+ (aq) + 12e- → 4Cr3+ (aq) + 7H2O
Add 14H+ to the oxidation half-reaction.
Step 8: Check the overall charge and balance it if necessary.
Oxidation: 6Ca(s) → 6Ca2+ (aq) + 12e-
Reduction: 2Cr2O7(2-) (aq) + 14H+ (aq) + 12e- → 4Cr3+ (aq) + 7H2O
The overall charge is balanced because there are 12 electrons transferred on both sides.
Step 9: Add both half-reactions together and cancel out the electrons.
6Ca(s) + 2Cr2O7(2-) (aq) + 14H+ (aq) → 6Ca2+ (aq) + 4Cr3+ (aq) + 7H2O
That's the balanced equation for the reaction in acid medium.
Now, let's move on to problem B.
B) Fe(CN)6(4-) (aq) + Ce(4+) (aq) → Ce(OH)3(s) + Fe(OH)3(s) + CO3(2-) (aq) + NO3(-)
The steps for balancing this reaction are similar to the previous one. Give it a try, and if you need any help, feel free to ask!