`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?A. Since `E^@` value for the cell will be positive, redox reaction is feasible.B. Since `E^@` value for the cell will be negative, redox reaction is not feasible.C. Ni cannot reduce `Ag^+` to Ag hence reaction is not feasible.D. Ag can reduce `Ni^(2+)` to Ni hence reaction is feasible.

`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . I

1.	`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?A. Since `E^@` value for the cell will be positive, redox reaction is feasible.B. Since `E^@` value for the cell will be negative, redox reaction is not feasible.C. Ni cannot reduce `Ag^+` to Ag hence reaction is not feasible.D. Ag can reduce `Ni^(2+)` to Ni hence reaction is feasible.
Answer» Correct Answer - A The cell reaction will be `Ni_((s))+2Ag_((aq))^+ to Ni_((aq))^(2+) + 2Ag_((s))` `E_"cell"^@=E_"cathode"-E_"anode"=0.80-(-0.25)= pm 1.05` V `DeltaG^@= -nFE_"cell"^@ As E_"cell"^@ = +ve`, `DeltaG^@=-ve` , hence reaction is feasible .

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