Question

question 8 (4 points)
balance the following reaction in a basic solution:
ag(s) → ag₂o(aq)

Explanation:

Step1: Split into half - reactions

Oxidation half - reaction: $Ag(s)\to Ag^{+}(aq)$
Reduction half - reaction: $O_{2}(g)\to H_{2}O(l)$ (since oxygen is involved in the formation of the oxide and in basic solution, we consider oxygen reacting to form water)

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

$2Ag(s)\to Ag_{2}^{2 + }(aq)$

Step3: Balance O and H in the reduction half - reaction in basic solution

$O_{2}(g)+2H_{2}O(l)+4e^{-}\to 4OH^{-}(aq)$

Step4: Balance the number of electrons in both half - reactions

The oxidation half - reaction: $2Ag(s)\to Ag_{2}^{2 + }(aq)+2e^{-}$ (multiply by 2 to get 4 electrons)
$4Ag(s)\to 2Ag_{2}^{2 + }(aq)+4e^{-}$

Step5: Combine the half - reactions

$4Ag(s)+O_{2}(g)+2H_{2}O(l)\to 2Ag_{2}O(s)+4OH^{-}(aq)$

Answer:

$4Ag(s)+O_{2}(g)+2H_{2}O(l)\to 2Ag_{2}O(s)+4OH^{-}(aq)$