Question

The voltage generated by the zinc concentration cell described by, zn(s)lzn2 (aq, o.100 Moll Zn2 (aq, M)Zn(so is 14.0 mV at 25 oC. Calculate the concentration of the Zn (aq) ion at the cathode. Number DO

Answer 1

Cell: ZnlZn2*(0.100 M) ll Zn2*(?)| Zn Since both electrodes of same material, this cell is a concentrati on cell and the standard cell potential will be zero Reduction half reaction: Zn2 (?)+ 2e Recduction half reaction: Zn2* (?)+ 2e Oxidation alf reaction : Cell reaction: -> Zn(s Zn(s) → Zn2+(0.100 M)+ 2e- Zn(s) + Zn2+(?) → Zn2+ (0.100 M) +Zn(s) Zn"(?) → Zn2+ (0.100 M) Standard cell potential, E EeaodeE anode = 0.00 V Hence, the emf of the cell under standard conditions is 0.00 V. Standard cell potential, E 0.00V No of moles of electrons transferredin thecell, n =2 Measured cell potential, Ecell-14.0 mV-0.0 140 V According to Nenst equation cell E .. pl 0.0591V İZn2. anode Zn thode 0.0591V 0.100 0.0 140 V = 0.00 V- lo Zn athode 0.0591V 0.100 [aTatlode =-0.0140 V 0.100 0.474 Zn athode 0.100 10047 0.336 Zn cathode [r lathode = 0.298 M Thus,the concentration of Zn- in cathode is 0.298 M