Question

Fill 10 km It w/ tup water and invert it in a 400 ml beaker billed with 300 - 350 ml of top water so that I the water remain in the test tube obtain a small piece of metallic calcium and rub it on an emery cloth until you can see a shiny metallic surten in one spot. place the metal gas, test the gas for Combusion by letting it out of the bealker, and with it mouth poinlect down turn it a a small angles as you bring it near a bunsen burnor. - Overall no ionic equation - predict the products formed in the beaker and the half reaction Method to balance it. (use the Family name of the metal as a hint in balancing).

Answer 1

According to the given information, Calcium has reacted with water, and it will produce calcium hydroxide and hydrogen gas.

Overall molecular equation is: Ca (s) + 2 H₂O (l) $\rightarrow$ Ca(OH)₂ (aq) + H₂ (g)

Overall ionic equation is: Ca + 2 H⁺ + 2 OH^-$\rightarrow$ Ca²⁺ + 2 OH^- + H₂

Balancing:

Now, as we know that calcium belongs to alkaline earth metal series (Group 2 of the periodic table) and can lose 2 electrons to achieve noble gas equation.

Eq. 1) Ca $\rightarrow$ Ca²⁺ + 2 e^-.

Water dissociates into hydrogen ion and hydroxide ion.

Eq. 2) 2 H₂O $\rightarrow$ 2 H⁺ + 2 OH^-

( This Equation is multiplied by 2 to balance the charge in the overall equation)

The H⁺ produced in eq. 2 can take up the electrons released by Ca in eq. 1.

Eq. 3) 2 H⁺ + 2 e^-$\rightarrow$ H₂

Since in equation 1, electrons are removed from the metal. Oxidation is taking place and this is the anodic half cell.

In equation 3, electrons are taken up by the hydrogen ions. Reduction is taking place and this is the cathodic half cell.

Add equation 1 and 3 to form net ionic equation.

Ca + 2 H⁺ + 2 e^-$\rightarrow$ Ca²⁺ + H₂ + 2e^-

Cancel out electron on both sides, we get

Ca + 2 H⁺$\rightarrow$ Ca²⁺ + H₂

In general, alkaline earth metal can be represented as M and its ion as M²⁺. So, you can replace all Ca in the above equations by M and Ca²⁺ by M²⁺ when you need to use family name.