Question

${M_2}O$ is the oxide of a metal ‘M’ which is above hydrogen in the activity series. ${M_2}O$ when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity.
(i) State the reaction taking place at the cathode.
(ii) Name the product at the anode

Answer 1

. At cathode, reduction process takes place and at anode, oxidation process takes place. Reduction is the gain of electrons whereas oxidation is the loss of electrons. Also, you must know that metals are good conductors because they readily lose their valence electrons.

Complete step by step answer:
We are given a metal oxide ${M_2}O$ of metal ‘M’ which is above the hydrogen atom in the activity series.
(i) Given that${M_2}O$ when dissolved in water forms the corresponding hydroxide. This chemical reaction can be represented as:
${M_2}O + {H_2}O \to MOH$

Here, MOH is the metal hydroxide of metal ‘M’.
Now, we all know metals have a tendency to easily lose their electrons. Now, we know that in the given oxide ${M_2}O$, valency of O is -2 and valency of metal ‘M’ is +1. Thus, metal ‘M’ has one electron in its valence shell. Hence, it will lose its one electron. When metal ‘M’ will lose its one electron, the reduction process occurs as shown below:
${M^ + } + {e^ - } \to M$
We know that reduction occurs at cathode therefore, the above shown reaction occurs at cathode.

(ii) Now, if metal ‘M’ is undergoing reduction, then oxygen atom will undergo oxidation. Oxidation occurs at anode. Thus, the product at the anode will be oxygen gas.

Note: Activity series, also called reactivity series, is a list of metals arranged in order of their reactivity with hydrogen ion sources, with most reactive at the top and least reactive at the bottom. The metal above the hydrogen atom in the reactivity series are good reducing agents, that means they undergo self-oxidation readily and reduce other elements or compounds readily.