Calculate the standard emf of the cell : `Cd|Cd^(2+)||Cu^(2+) | Cu`are `0.34` V and `-0 .40` volts respectively. Predict the feasibility of the cell reaction .

Explanation

Cell reactions :
Oxidation at anode :
`Cd(s) to Cd^(2+)(aq) + 2e^(-) (E_("ox")^(0))_(cd|cd^(2+)) = 0.4 V`
Reduction at cathode :
`Cu^(2+)(aq) + 2e^(-) to Cu(s) (E_("red")^(0))_(Cu^(2+)|Cu) = 0.34 V`
`Cd(s) + 2e^(-) to Cd^(2+) (aq) + Cu(s)`
`E_("cell")^(0) = (E_("ox")^(0)) + (E_("red")^(0))_("cathode")`
` = 0.4 + 0.34`
` = 0.74 V`
emf is positive , so `Delta G` is negative , the reaction is feasible.