1.

Standardelectrode potential of ell H_(2)|H^(+)||Ag^(+)|Ag is(Ag^(+)//Ag)^(@)=0.80 V

Answer»

0.8 V
`-0.8 V`
`-1.2 V`
`1.2 v`

Solution :`E_(cell) =E_(cathode)-E_(anode)`
Given `E_(AG^(+)//Ag)^(@)=0.80 V H_(2) |H^(+)||Ag^(+)|Ag`
`THEREFORE`HYDROGEN is anode and silver is cathode
`E_(cell)=E_(c )-E_(A)`
=0.80 -0
=0.80 V


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