balance the redox reaction

NO2^- + Ag^+ = NO3^- + Ag

Ag^+ + e ==> Ag

H2O+ NO2^- ==> NO3^- + 2e + 2H^+

Multiply equation 1 by 2 and equation 2 by 1 then add the two for the complete equation.

wouldnt h2o be on the other side because you orignally have 2h20 on the products and one h20 in the reactants so cancelling them out would make one h20 on the products side

sorry this is in basic solution

To balance the redox reaction, follow these steps:

Step 1: Identify the oxidation states of each element in the reaction.
In this reaction, nitrogen (N) has an oxidation state of +4 in NO2^- and +5 in NO3^-. Silver (Ag) has an oxidation state of +1 in Ag^+.

Step 2: Separate the reaction into two half-reactions for the oxidation and reduction processes.
Oxidation half-reaction: NO2^- -> NO3^-
Reduction half-reaction: Ag^+ -> Ag

Step 3: Balance the atoms other than oxygen and hydrogen in each half-reaction.
Oxidation half-reaction: 2NO2^- -> 2NO3^-
Reduction half-reaction: Ag^+ -> Ag

Step 4: Balance the oxygen atoms in each half-reaction by adding water (H2O) molecules.
Oxidation half-reaction: 2NO2^- + H2O -> 2NO3^- + 2e^-
Reduction half-reaction: Ag^+ -> Ag + H2O

Step 5: Balance the hydrogen atoms in each half-reaction by adding hydrogen ions (H+).
Oxidation half-reaction: 2NO2^- + H2O -> 2NO3^- + 2e^-
Reduction half-reaction: Ag^+ + H2O -> Ag + H^+

Step 6: Balance the charges in each half-reaction by adding electrons (e^-).
Oxidation half-reaction: 2NO2^- + H2O -> 2NO3^- + 2e^-
Reduction half-reaction: Ag^+ + H2O + e^- -> Ag + H^+

Step 7: Multiply the half-reactions by a whole number to make the number of electrons equal in both half-reactions.
Oxidation half-reaction: 2NO2^- + H2O -> 2NO3^- + 2e^-
Reduction half-reaction: 2Ag^+ + 2H2O + 2e^- -> 2Ag + 2H^+

Step 8: Add the two balanced half-reactions together.
2NO2^- + H2O -> 2NO3^- + 2e^-
2Ag^+ + 2H2O + 2e^- -> 2Ag + 2H^+
---------------------------
2NO2^- + 2Ag^+ -> 2NO3^- + 2Ag

The balanced redox reaction is 2NO2^- + 2Ag^+ -> 2NO3^- + 2Ag.