chapter 5 electrochemistry class 12 chemistry textbook solution

4. Answer the following :

vi. Calculate emf of the following cell at 250C. Zn (s) |Zn2⊕(0.08M)||Cr3⊕(0.1M)|Cr E0 
Zn = – 0.76 V, E0 Cr = – 0.74 V (0.0327 V)

Answer:-

Given: EZn=0.76V,ECr=0.74V To find: Emf of the cell (Ecell) Formulae: 1) Ecell=EcathodeEanode 2) Ecell=Ecell0.0592Vnlog10[ProductReactant] Calculation: 3Zn(s)2Cr(0.1M)3+3Zn(0.08M)2++Cr(s)(overall reaction) Using formula (1): Ecell=EcathodeEanode Ecell=ECrEZn =0.74V(0.76V)=0.02V Using formula (2): The cell potential is given by Ecell=Ecell0.0592V6log10(0.0830.12) =0.02V+0.0127=0.0327V The emf of the cell is 0.0327V.

chapter 5 elctrochemistry textbook solution page 118