Consider the unbalanced redox reaction:
Cr2 O7 2-(aq) + Cu(s) ----> Cr3+(aq) + Cu2+
Balance the equation in acidic solution and determine how much of a 0.850 M K2 Cr2 O7 solution is required to completely dissolve 5.25 g of Cu.
How much of this can you do on your own? Where are you stuck?
on chem chapter 16
To balance the redox equation in acidic solution, follow these steps:
Step 1: Write the unbalanced equation:
Cr2 O7 2-(aq) + Cu(s) ----> Cr3+(aq) + Cu2+
Step 2: Split the equation into two half-reactions:
Half-reaction 1: Cr2 O7 2-(aq) ----> Cr3+(aq)
Half-reaction 2: Cu(s) ----> Cu2+
Step 3: Balance the atoms in each half-reaction:
For the first half-reaction, we have 2 Cr on both sides, 7 O on the reactant side, and 6 O on the product side. To balance oxygen, you need to add H2 O to the reactant side:
Cr2 O7 2-(aq) + 4 H+(aq) ----> 2 Cr3+(aq) + 7 H2 O(l)
For the second half-reaction, we already have no hydrogen or oxygen atoms, so it is already balanced:
Cu(s) ----> Cu2+
Step 4: Balance the charges in each half-reaction:
In the first half-reaction, we have a charge of -2 on the reactant side (since we have Cr2 O7 2-) and a charge of +6 on the product side (since we have Cr3+). To balance the charges, add electrons (e-) to the product side:
Cr2 O7 2-(aq) + 14 H+(aq) + 6 e- ----> 2 Cr3+(aq) + 7 H2 O(l)
In the second half-reaction, we have a charge of 0 on both sides, so no adjustment is needed.
Step 5: Balance the number of electrons transferred by multiplying the half-reactions by appropriate numbers so that the electrons cancel out when added together:
In this case, multiply the first half-reaction by 6 and the second half-reaction by 3 to equalize the number of electrons transferred:
6(Cr2 O7 2-(aq) + 14 H+(aq) + 6 e-) ----> 12 Cr3+(aq) + 42 H2 O(l)
3(Cu(s) ----> Cu2+)
Step 6: Add the two half-reactions together to get the balanced equation:
6 Cr2 O7 2-(aq) + 42 H+(aq) + 18 e- + 3 Cu(s) ----> 12 Cr3+(aq) + 42 H2 O(l) + 3 Cu2+
Finally, the balanced equation in acidic solution is:
3 Cu(s) + 6 H+(aq) + 6 Cr2 O7 2-(aq) ----> 3 Cu2+(aq) + 6 Cr3+(aq) + 3 H2 O(l)
To determine how much of a 0.850 M K2 Cr2 O7 solution is required to completely dissolve 5.25 g of Cu, we need to perform a stoichiometric calculation.
Step 1: Calculate the moles of Cu:
molar mass of Cu = 63.55 g/mol
moles of Cu = mass/molar mass = 5.25 g / 63.55 g/mol = 0.0826 mol
Step 2: Use the balanced equation to find the mole ratio between Cu and K2 Cr2 O7:
From the balanced equation, we can see that the mole ratio of Cu to K2 Cr2 O7 is 3:6, or 1:2.
Step 3: Calculate the moles of K2 Cr2 O7 required:
moles of K2 Cr2 O7 = 2 * moles of Cu = 2 * 0.0826 mol = 0.165 mol
Step 4: Calculate the volume of the K2 Cr2 O7 solution using the molarity:
volume (in liters) = moles / molarity = 0.165 mol / 0.850 mol/L = 0.194 L
Therefore, you would need approximately 0.194 liters (or 194 mL) of the 0.850 M K2 Cr2 O7 solution to completely dissolve 5.25 g of Cu.
To balance the redox reaction in acidic solution, follow these steps:
Step 1: Assign oxidation numbers to each atom in the reaction.
In this reaction:
- The oxidation number of Cr in Cr2O7^2- is +6.
- The oxidation number of O in Cr2O7^2- is -2.
- The oxidation number of Cu in Cu(s) is 0.
- The oxidation number of Cr in Cr^3+ is +3.
- The oxidation number of Cu in Cu^2+ is +2.
Step 2: Identify the atoms undergoing oxidation and reduction.
- Cr is reduced from +6 to +3, so it is being reduced (undergoing reduction).
- Cu is oxidized from 0 to +2, so it is being oxidized (undergoing oxidation).
Step 3: Write separate half-reactions for oxidation and reduction.
Oxidation half-reaction:
Cu(s) → Cu^2+(aq) + 2e-
Reduction half-reaction:
Cr2O7^2-(aq) + 14H+ + 6e- → 2Cr^3+(aq) + 7H2O(l)
Step 4: Balance each half-reaction individually.
Since we have 6 electrons in the reduction half-reaction, we need to multiply the oxidation half-reaction by 6 to balance the electrons.
6Cu(s) → 6Cu^2+(aq) + 12e-
Cr2O7^2-(aq) + 14H+ + 6e- → 2Cr^3+(aq) + 7H2O(l)
Step 5: Combine the half-reactions and eliminate any spectator ions.
To eliminate the electrons, multiply the oxidation half-reaction by the number of electrons in the reduction half-reaction (12) and add both reactions together.
6Cu(s) + Cr2O7^2-(aq) + 14H+ → 6Cu^2+(aq) + 2Cr^3+(aq) + 7H2O(l)
Step 6: Balance the equation by adding water and H+ ions as needed.
In this case, we need to add 14 H2O to the left side:
6Cu(s) + Cr2O7^2-(aq) + 14H+ → 6Cu^2+(aq) + 2Cr^3+(aq) + 7H2O(l)
Now the reaction is balanced in acidic solution.
To determine how much of a 0.850 M K2Cr2O7 solution is required to completely dissolve 5.25 g of Cu, we need to determine the stoichiometry of the reaction.
Step 1: Calculate the number of moles of Cu.
Molar mass of Cu = 63.55 g/mol
Moles of Cu = mass of Cu / molar mass of Cu = 5.25 g / 63.55 g/mol
Step 2: Use the balanced equation to determine the stoichiometry.
From the balanced equation, we can see that the ratio of Cu to Cr2O7^2- is 6:1. Thus, 6 moles of Cu react with 1 mole of Cr2O7^2-.
Step 3: Calculate the number of moles of K2Cr2O7 required.
Moles of K2Cr2O7 = (moles of Cu) / 6
Step 4: Calculate the volume of the K2Cr2O7 solution.
Volume (L) of K2Cr2O7 solution = (moles of K2Cr2O7) / (concentration of K2Cr2O7 in Molarity)
In this case, the concentration of K2Cr2O7 is given as 0.850 M.
Volume (L) of K2Cr2O7 solution = (moles of K2Cr2O7) / 0.850 M
By following these steps, you can calculate the volume of the K2Cr2O7 solution required to completely dissolve 5.25 g of Cu.