in a particular redox reaction, Cr is oxidized to CrO42- and Ag+ is reduced to Ag. Complete and balance the equation for this reaction in acidic solution.

To balance the redox equation for this particular reaction in acidic solution, you can follow the steps below:

Step 1: Write the unbalanced equation for the reaction.
Cr + Ag+ → CrO42- + Ag

Step 2: Assign oxidation states to each element in the reaction.
Cr is initially 0 and becomes +6 in CrO42-.
Ag+ is initially +1 and becomes 0 as Ag.
Therefore, Cr is oxidized, and Ag+ is reduced.

Step 3: Separate the equation into two half-reactions -- oxidation and reduction.
Oxidation Half-Reaction: Cr → CrO42-
Reduction Half-Reaction: Ag+ → Ag

Step 4: Balance the atoms involved in each half-reaction, excluding H and O.
Oxidation Half-Reaction: Cr → CrO42-
To balance the Cr atoms, add 4 Cr on the right side.
Cr → 4 CrO42-

Reduction Half-Reaction: Ag+ → Ag
The Ag atoms are already balanced.

Step 5: Balance the oxygen atoms by adding H2O molecules to the side that needs oxygen.
Oxidation Half-Reaction: Cr → 4 CrO42-
To balance the oxygen atoms, add 7 H2O on the left side.
Cr + 7 H2O → 4 CrO42-

Reduction Half-Reaction: Ag+ → Ag
The number of oxygen atoms remains unchanged.

Step 6: Balance the hydrogen atoms by adding H+ ions to the side that needs hydrogen.
Oxidation Half-Reaction: Cr + 7 H2O → 4 CrO42-
To balance the hydrogen atoms, add 14 H+ on the right side.
Cr + 7 H2O → 4 CrO42- + 14 H+

Reduction Half-Reaction: Ag+ → Ag
To balance the hydrogen atoms, add 2 H+ on the left side.
2 H+ + Ag+ → Ag

Step 7: Balance the charge in each half-reaction using electrons (e-).
Oxidation Half-Reaction: Cr + 7 H2O → 4 CrO42- + 14 H+ + 6e-
Reduction Half-Reaction: 2 H+ + Ag+ → Ag

Step 8: Multiply each half-reaction by a factor that equalizes the number of transferred electrons.
Oxidation Half-Reaction: 3(Cr + 7 H2O → 4 CrO42- + 14 H+ + 6e-)
6(Cr + 7 H2O → 4 CrO42- + 14 H+ + 6e-) will cancel out the 6e-.
Reduction Half-Reaction: 6(2 H+ + Ag+ → Ag) will cancel out the 6e-.

Step 9: Combine the half-reactions to generate the balanced redox equation.
6(Cr + 7 H2O → 4 CrO42- + 14 H+ + 6e-) + 3(Ag+ + 2 H+ → Ag)
6Cr + 42 H2O → 24 CrO42- + 84 H+ + 36e- + 3Ag+ + 6 H+ → 3Ag

Step 10: Simplify and eliminate any redundant terms.
6Cr + 42 H2O + 3Ag+ → 24 CrO42- + 90 H+ + 3Ag

So, the balanced redox equation for Cr oxidation and Ag+ reduction in acidic solution is:
6Cr + 42 H2O + 3Ag+ → 24 CrO42- + 90 H+ + 3Ag

To balance the redox reaction in acidic solution, follow these steps:

Step 1: Write the unbalanced equation.
Cr + Ag+ -> CrO42- + Ag

Step 2: Separate the reaction into two half-reactions - oxidation and reduction.
Oxidation half-reaction: Cr -> CrO42- (Cr is oxidized)
Reduction half-reaction: Ag+ -> Ag (Ag+ is reduced)

Step 3: Balance the atoms in each half-reaction.
Oxidation half-reaction: Cr -> CrO42-
To balance the number of Cr atoms, add 7 H2O to the right side:
Cr -> CrO42- + 7H2O

Reduction half-reaction: Ag+ -> Ag
This is already balanced.

Step 4: Balance the charge in each half-reaction.
Oxidation half-reaction: Cr -> CrO42-
To balance the charge, add 6 electrons (e-) to the left side:
Cr + 6e- -> CrO42-

Reduction half-reaction: Ag+ -> Ag
No additional electrons are needed as the charges are already balanced.

Step 5: Equalize the number of electrons in both half-reactions.
To equalize the number of electrons, multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1:
6(Cr + 6e- -> CrO42-) + Ag+ -> 6Ag

Step 6: Combine the half-reactions.
Add the two half-reactions together, canceling out common species on both sides of the reaction:
6Cr + 42H2O + 6Ag+ -> 6CrO42- + 6Ag + 7H2O

Step 7: Simplify the equation.
Simplify the equation by eliminating the duplicated water molecules:
6Cr + 14H2O + 6Ag+ -> 6CrO42- + 6Ag

Therefore, the balanced redox equation for the reaction in acidic solution is:
6Cr + 14H2O + 6Ag+ -> 6CrO42- + 6Ag

Here is a site that shows how to do all of this. If you still have trouble, be specific and we can help you through the next step. http://www.chemteam.info/Redox/Redox.html