calculate the emf of the following cell is found to be 0.20V at 298K cd|cd2+|| Ni2+(2.0M)|Ni what is the molar concentration of cd2+ ions in the solution.

Dear Student,

The cell reaction isCd(s) + Ni2+(aq)  Cd2+(aq) + Ni(s)and, Ecell=E°cell+0.0591nlog[Ni2+][Cd2+][Ni2+]=2n=2Here, E°cell=E°Ni2+/Ni-E°Cd2+/Cd=-0.25-(-0.40)=0.15 VEcell=0.20 VTherefore, 0.20=0.15 +0.05912log2[Cd2+]     0.05=0.05912log 2[Cd2+]or, log2[Cd2+]=1.69or, log 2- log[Cd2+] =1.69or, log [Cd2+]=log 2-1.69 =-1.39or. [Cd2+]=Antilog(-1.39)=0.0407 M 

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