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)
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!
A redox reaction involves loss and gain of electrons.
A redox reaction is one in which a specie accepts electrons and another specie donates electrons. The number of electrons lost must be balanced by the number of electrons gained in the reaction.
For the redox reaction equation in which Cr is oxidized to CrO4^- and Ag^+ is reduced to Ag, the balanced reaction equation in alkali is;
[tex]6Ag^+ + Cr + 8OH^- ---> 6Ag + CrO4^{2-} + 4H2O[/tex]
Learn more: https://brainly.com/question/2510654
