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The reaction taking place in the cell Pg...

The reaction taking place in the cell `Pg|H_(2)(g)|HCl(1.0M)|AgCl|Ag` is 1 atm

A

`AgCl+(1//2)H_(2)toAg+H^(+)+Cl^(-)`

B

`Ag+H^(+)+Cl^(-)toAgCl+(1//2)H_(2)`

C

`2Ag^(+)+H_(2)to2Ag+2H^(+)`

D

`2Ag+2H^(+)to2Ag^(+)+H_(2)`.

Text Solution

Verified by Experts

The correct Answer is:
A
LHS electrode (oxidation half reaction)
`(1)/(2)H_(2)toH^(+)+e^(-)`
RHS electroe (reduction half reaction)
`AgCl+e^(-)toAg+Cl^(-)`
Adding the two half ractions, we get
`AgCl+e^(-)toAg+H^(+)+Cl^(-)`
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