For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Ag + is reduced to Ag . Complete and balance the equation for this reaction in basic solution. The phases are optional. balanced reaction: Cr + Ag^{+} -> CrO_{4}^{2}^{-} + Ag Cr + Ag + ⟶ CrO 2 − 4 + Ag

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺   →   Ag      

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ →  Ag   (1)

Now, we need to balance the half-reaction of Cr:

Cr   →  CrO₄²⁻  

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:          

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻  

The hydrogen atoms will be balanced using H₂O molecules:  

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻ + H₂O    

The balanced equation is:

Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O   (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ →  Ag)  

Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O                  

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

the answer is 0.42 is that is an option. this question provides no info