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The anodic solution of standard " Al-Ag" voltaic dilute by 100 times. The potential of the cell is increased by '0.01x' volts what is x ?

Answer»


Solution :`Al + 3 AG^(oplus) to Al^(+3) + 3Ag , E=E^(0) - (0.0591)/(3) LOG""([Al^(+3)])/([Ag^(+1)]^(3)) , E_1 = E^(0) - (0.06)/(3) log""(C)/([Ag^(+)]^(3))`
` in_2 = E^(0) - (0.06)/(3) log ""(((C)/(100)))/([Ag^(+1)]^(3)), DeltaE = (in_2 - in_1) = - 0.02 log [(C)/(100) xx (1)/([Ag^(+)]^(3)) xx ([Ag^(+)]^(3))/(e^(-))]`
`=-0.02 log 10^(-2) = - 0.02 xx (-2) log_(10)^(10) = 0.04 = 0.01 X = 0.04 , x =4 `


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