3. -133.34 points FSUGenChem2L1 9.PRE.003. My Notes o Ask Your Teacher AG° and E

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3. -133.34 points FSUGenChem2L1 9.PRE.003. My Notes o Ask Your Teacher AG° and EP can be said to measure the same thing, and are convertible by the equation AG° = -FE cell where n is the total number of moles of electrons being transferred, and F is the Faraday constant 9.64853415X104 C/mol. The free energy (AG°) of a spontaneous reaction is always negative. For each of the electrochemical cells below, calculate the free energy of the system and state whether the reaction is spontaneous or non-spontaneous as written based on the cathode and anode assignment given. (Use the table of Standard Electrode Potentials.) (a) The cathode is Ni(II) and the anode is Sn(II). free energy spontaneity ---Select-4 (b) The cathode is Ag(1) and the anode is Pb(II). free energy spontaneity --Select-- (c) The cathode is Zn(II) and the anode is Cd(II). free energy Z spontaneity ---Select--- k) Supporting Materials Il periodic Table al Constants and Supplemental Factors Data Submit Answer Save Progress | Practice Another Version

Explanation / Answer

At Cathode oxidation occurs. Ni loose electrons Ni------->Ni+2+2e-, Eo=2.3V (1)

At anode reduction occurs. Sn+2+2e- ------->Sn, Eo = -0.14V (2)

Eq.1+Eq.2 gives the overall reaction Ni+Sn+2------->Ni+2+Sn, Eo= 2.3-0.14= 2.16V

deltaG= -nFE, n = no of electrons exchanged= 2, F= 96500 C/mole and E= 2.16V

deltaG= -2*96500*2.16= -416880 J/mole=-416.88 Kj/mole

2. at cathode, Ag(s) -------> Ag+ (aq) + e- ,         E =-.80V   (1)

Pb+2+2e- --------->Pb, Eo= -0.13V   (2)

multiplying Eq.1 with 2 and addition with Eq.2 gives

2Ag(s)+ Pb+2 --------->2Ag+ + Pb, Eo= -0.8-0.13=-0.93V, no of electrons exchanged= 2

deltaG= -2*96500*(-0.93)=179490 J/mole =179.49 Kj/mole

3. Zn----->Zn+2+2e-, Eo= 0.76V (1) Cd 2+   +    2e-  <----->   Cd         E = =-0.40 V   (2)

Eq.1+Eq.2 gives Zn+Cd+2-------->Zn+2+Cd, Eo= 0.76-0.40=0.36V

deltaG=-nFE= -1*96500*0.36 J=-34740j=-34.740 KJ

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