1.

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

Answer»

`AGCL+(1//2)H_(2)toAg+H^(+)+Cl^(-)`
`Ag+H^(+)+Cl^(-)toAgCl+(1//2)H_(2)`
`2Ag^(+)+H_(2)to2Ag+2H^(+)`
`2Ag+2H^(+)to2Ag^(+)+H_(2)`.

SOLUTION :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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