Solve Share with your friends Share 1 Vartika Jain answered this Dear Student, Now, E°Zn = -0.76 . The negative sign shows that Zn electrode undergoes oxidation i.e. it serves as anode. Hydrigen gas electrode acts as anode. Therefore, the cell will be represented as :Zn(s)|Zn2+(aq)(0.24 M) || H+(aq)(1.6 M)|H2(g) 1.8 atmThe half reactions are Zn→ Zn2++2e-H2+2e- → 2H+n=2Now, E°cell= E°cathode-E°anode= 0-(-0.76)=0.76 VEcell=E°cell-0.0591nlog[Zn2+]pH2[H+]2 =0.76-0.05912log0.24×1.8(1.6)2 =0.76+0.02 =0.78 V 0 View Full Answer
