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Standard VII

Chemistry

Chemical Formula of Ionic Compounds

M is a metal ...

Question

$M$ is a metal above hydrogen in the activity series and its oxide has the formula $M_{2} O$ . This oxide when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity. In the above context answer the following:
State the reaction taking place at the cathode.

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Solution

The valency of O is -2. As 2 atoms of M are combined in $M_{2} O$ the valency of M is +1. so, at cathode M will lose its 1 valence electron and from neutral atom. $M^{+} + e^{-} \to M$ . Hence, it will undergo reduction reaction at the cathode.

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(a) The observation seen when fused lead bromide is electrolysed is :

(i) A silver grey deposit at an anode and a reddish brown deposit at the cathode.

(ii) A silver grey deposit at the cathode and a reddish brown deposit at the anode.

(iii) A silver grey deposit at the cathode and reddish brown fumes at the anode.

(iv) Silver grey fumes at anode reddish brown fumes at the cathode.

(b) During electroplating an article with silver, the electrolyte used is :

(i) Silver nitrate solution

(ii) Silver cyanide solution

(iii) Sodium argentocyanide solution

(iv) Nickel sulphate solution.

(d) State your observation: At the cathode when acidified aqueous copper sulphate solution is electrolysed with copper electrodes.

(e) State which electrode: anode or cathode is the oxidising electrode. Give a reason for the same.

(f) Name the king of particle present in :

(i) Sodium hydroxide soln.

(ii) Carbonic acid

(iii) Sugar soln.

(g) $M_{2} O$ 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.

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M is a metal above hydrogen in the activity series and its oxide has the formula M2O. This oxide when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity. In the above context answer the following:State the reaction taking place at the cathode.

The valency of O is -2. As 2 atoms of M are combined in M2O the valency of M is +1. so, at cathode M will lose its 1 valence electron and from neutral atom. M++e−→M. Hence, it will undergo reduction reaction at the cathode.

$M$ is a metal above hydrogen in the activity series and its oxide has the formula $M_{2} O$ . This oxide when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity. In the above context answer the following:
State the reaction taking place at the cathode.

The valency of O is -2. As 2 atoms of M are combined in $M_{2} O$ the valency of M is +1. so, at cathode M will lose its 1 valence electron and from neutral atom. $M^{+} + e^{-} \to M$ . Hence, it will undergo reduction reaction at the cathode.