Question

In the above cell cathodic cell reaction and number of electrons involved are:

• A. $2A{g}^{+}+2e^{-}\to 2Ag,2e^{-}$
• B. $A{g}^{+}+e^{-}\to Ag,1e^{-}$
• C. $2H^{+}+2e^{-}\to H_2,2e^{-}$
• D. $H^{+}+e^{-}\to \frac{1}{2}{H}_2,e^{-}$

Answer 1

The correct option is A $2A{g}^{+}+2e^{-}\to 2Ag,2e^{-}$

$2A{g}^{+}\left(aq\right)+H_2\left(g\right)\to 2Ag\left(s\right)+2H^{+}\left(aq\right),$

This is reduction cell of silver ion with respect to hydrogen so for this cell,

At cathode:
$2A{g}^{+}+2e\to 2Ag$.

At anode:
$H_2\to 2H^{+}+2e$

so electrons involved in this reaction are 2.