Given the unbalanced equation Ca(s) + Ag+(aq) → Ca2+(aq) + Ag(s), for a reaction, what is being reduced?

Ag^+ is reduced to Ag^0

Explanation:

Based on the equation

Ca+Ag^+=Ca^2+ + Ag

Ca^0 is increased to Ca^2+

And Ag^+ is reduced to Ag^0

Ag is reduced in the reaction.

Explanation: From the balanced equation: Ca + Ag+ => Ca2+ + Ag

For an element to be reduced there must be a decrease in oxidation number. For Ca in the reactant it is an independent atom so its oxidation number is 0. The on the product the oxidation number of Ca increases to 2+. So Ca is oxidized.

Ag in the reactant has an oxidation of +1 and decreases on the product side where its oxidation is 0. So Ag is reduced in the reaction since there is a decrease in oxidation number.

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.