Use oxidation numbers to find the oxidant and the reductant!
$NO_3^-$ $(o.n.(N)=5)$ is the oxidant which is reduced to $NO$ $(o.n.(N)=2)$. $Ag$ $(o.n.(Ag)=0)$ is the reductant which is oxidised to $Ag^{+}$ $(o.n.(Ag)=1)$.
How many electrons are exchanged by the oxidant and the reductant?
$NO_3^-$ catches 3 $e^-$. $Ag$ looses 1 électron
Write the partial oxidation and reduction reactions in an acid solution.
$NO_3^- $ $+$ $3e^-$ $+$ $4H^+$ $\longrightarrow$ $NO $ $+$ $2H_2O$ $Ag$ $-$ $e^-$ $\longrightarrow$ $Ag^+$
Balance by the number of exchanged electrons and write the redox reaction!
$NO_3^-$ $+$ $3e^-$ $+$ $4H^+$ $\longrightarrow$ $NO$ $+$ $2H_2O$
$3Ag$ $-$ $3e^-$ $\longrightarrow$ $3Ag^+$
$NO_3^-$ $+$ $3Ag$ $+$ $4H^+$ $\longrightarrow$ $3Ag^+$ $+$ $NO$ $+$ $2H_2O$
Complete for the ions which are not involved et annotate!
$3Ag$ $+$ $4H^+$ $+$ $4NO_3^-$ $\longrightarrow$ $3Ag^+$ $+$ $3NO_3^-$ $+$ $NO$ $+$ $2H_2O$ Nitric acid and silver produce a solution of silver(I) nitrate, nitrogen monoxide (colourless) and water.
Recall the reaction of nitrogen monoxide with the oxigen present in air! What is this type of reaction?
$2NO$(colourless gas)$+$ $O_2$ $\longrightarrow$ $2NO_2$ (brown gas) It is a redox reaction, because the o.n. of N goes from 2 to 4 at the cost of the oxigen atoms of the molecule $O_2$