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Cell notation of a galvanic cell is represented as Ni(s)/Ni+2(M1)//Cl-(M2)//12Cl2(1atm), Pt(s). Emf of the cell is minimum when the values of M1 and M2 are

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a
M1=M2=0.01 M
b
M1=1 M ; M2=1 M ;
c
M1= 0.1 M ; M2= 0.01 M ;
d
M1= 0.01 M ; M2= 0.1 M ;

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detailed solution

Correct option is B

Anode (LHE) Ni(s)→ Ni+2(M1)+2e;Cathode (RHE) Cl2(1atm)+2e→ 2Cl-(M2)Net cell reaction : Ni(s)+Cl2(1atm)→ Ni+2(M1)+2Cl-(M2)Ecell=Ecell0-0.05912logNi+2Cl-2 ∴Ecell∝-logNi+2Cl-2Thus lesser the product of Ni+2Cl-2 higher will be the Ecell;The product of  Ni+2Cl-2 in each case is(1) option ….10-6(2) option ….1(3) option ….10-5(4) option ….10-4Minmum emf of the cell is when Ni+2=1 M and Cl- =1 M

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