Answer:
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)
<u>Cr + 8OH⁻ → CrO₄²⁻ + 6e⁻ + 4H₂O</u>
6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O
Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.
I hope it helps you!
