Question

Review I Constants I Periodic Table The equilibrium constant, K, for a redox reaction is related to the standard potential, E°, by the equation Standard reduction potentials nFE RT In K E° (V) Reduction half-reaction Agt(aq)eAg(s) Cu2+(aq)2eCu(s) 0.80 where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol - K) , and T is the Kelvin temperature. 0.34 Sn4t (aq)4eSn(s) 0.15 2H (aq)2e H2(g) Ni2+ (aq)2e->Ni(s) 0 -0.26 Fe2t(aq)2eFe(s) -0.45 Zn2 (aq)2eZn(s) -0.76 Al3 (aq)3e->Al(s) -1.66 Mg2+ (aq)2e Mg(s) -2.37

Answer 1

Past A: Fe(s) + Ni2+ (aq) → Feat (09) + Ni(s) € Given: So, Fe 2+ (aq) + 2 Fe (5) Fe(s) : E = -0.45 v Fe2+ (aq) tzé : Fox +0.45V ** For the reverse reaction | Anode cell reaction 4 Oxidation Reduction Cathode cell reaction And, Ni2+ (aq) + Z é and I sign of to get reversed. jole, Ereduction - - Eoscidation vice - Ni(s) ; Ered= – 0.26 V versa - Overall cell reaction: Fe(s) + Ni2+ (aq) Fe2+ (aq) + Nics) Standard Overall a potential : Ecell = E Cathode - Eanode = -0.26 V-(-0.45 v) = + 0.19 V ** For calculating Eien, - which is a reduction potential of overall cell reaction, reduchom potentials of both anode and cathode (i.e. Ered) are used. Now, to calculate the equilibrium constant, we have MFE°, 2.X 96500 C molex Olav 8.314 J. molek x298K ln K at T= 2988: RT (25°C) 7 enk - 14.8 note: CV - J = 14-8 = 2.67x106) ( -0.190) and other values are given

* Part B x(s) + yo can) → x®(aq) + Y () Gives, K - 6.96 x103 We have enk- NEE - RT Here , n=1 [ *(5) xaq) + ] Lytlan) té → Y2 T= Standard temperature = 25°C = 298 K R= 8.314 J. mole. K' - 96500 c so, E = lok x RT 8.314 J. molekx 298k or 8 = en (6.96 810 3 x 1 x 96500 C mole! or, = (-4.968) x 0.0257 V [: 2-v] & Hence, Ecell = -0.128 v , Expressed to three significant figures.