7. balance the following redox reaction that takes place in acidic solution: ag₂s + no₃⁻ → ag⁺ + s + no

Explanation:

Step1: Identify oxidation - reduction half - reactions

Oxidation: $Ag_2S\rightarrow 2Ag^++S$ Reduction: $NO_3^-\rightarrow NO$

Step2: Balance atoms other than O and H in each half - reaction

The oxidation half - reaction $Ag_2S\rightarrow 2Ag^++S$ is already balanced for atoms other than O and H. For the reduction half - reaction $NO_3^-\rightarrow NO$, N is already balanced.

Step3: Balance O atoms by adding $H_2O$

Oxidation: No O atoms, so no change. Reduction: $NO_3^-\rightarrow NO + 2H_2O$

Step4: Balance H atoms by adding $H^+$

Oxidation: No H atoms, so no change. Reduction: $4H^++NO_3^-\rightarrow NO + 2H_2O$

Step5: Balance charge by adding electrons

Oxidation: $Ag_2S\rightarrow 2Ag^++S+2e^-$ Reduction: $3e^- + 4H^++NO_3^-\rightarrow NO + 2H_2O$

Step6: Make the number of electrons equal in both half - reactions

Multiply the oxidation half - reaction by 3 and the reduction half - reaction by 2. Oxidation: $3Ag_2S\rightarrow 6Ag^++3S + 6e^-$ Reduction: $6e^-+8H^++2NO_3^-\rightarrow 2NO + 4H_2O$

Step7: Combine the two half - reactions

$3Ag_2S+8H^++2NO_3^-\rightarrow 6Ag^++3S + 2NO + 4H_2O$

Answer:

$3Ag_2S+8H^++2NO_3^-\rightarrow 6Ag^++3S + 2NO + 4H_2O$