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 maximum when the values of M1 and M2 are
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
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 ;
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
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=0.01 M and Cl- =0.01 M