Question

Consider a voltaic (galvanic) cell with the following metal electrodes. Identify which metal is the cathode and which is the anode, and calculate the cell potential. (Use the table of Standard Electrode Potentials.) (a) Ca(II) and Sc(III) Cathode: . Ca(II) Sc(III) Anode: Ca(II) Sc(III) Ecell = 0.0591 x V (b) Pb(II) and In(III) Cathode: . Pb(II) In(III) Anode: Pb(II) In(III) Ecell -

Answer 1

Cd2+ + 2 e− ---->   Cd(s)   −0.40 V .At cathode
Sc(s)---->   Sc3+ + 3 e   −2.077 V .At anode

E0cell = E0cathod-Eoanod = -0.4 -[-2.077] v = 1.677 V

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   In(s)--->In3+ + 3 e−       −0.34 V .at anode

   Pb2+ + 2 e−   --->   Pb(s)   −0.126 . At cathode

E0cell = E0cathod-Eoanod = [-0.126 V]-[-0.34V ] = 0.214 V

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   Zr(s)--->Zr4+ + 4 e−       −1.45v .At anode
Ni2+ + 2 e−--->   Ni(s)   −0.25 V At catode
E0cell = E0cathod-Eoanode =−0.25 V -[−1.45v ] 1.2 V
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The cell with higer reductionpotential will undergo reduction,other cell wii under go oxidation
So half reactions will be

Mg --> Mg2+ +2e- . Oxidation.at anode
Au3+ +3e----> Au(s).Reduction..at cathode
balance # of electrons by multipling first reaction with 3 and 2nd reaction with 2
3Mg --> 3Mg2+ +6e- . Oxidation.at anode
2Au3+ +6e----> 2Au(s).Reduction..at cathode

Add both reaction we get overall reaction

3Mg(s) + 2Au3+(aq) --> 2Au(s) + 3Mg2+(aq)
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Co2+ + 2 e− -->   Co(s)   −0.28 cathode

   Pb(s)---->Pb2+ + 2 e−       −0.126 anode

E0cell = E0cathod-Eoanode =- 0.28 v -{-0.126 V] = -0.154 V
number of electrons involved n = 2
dG0= -nfE = -[2*-0.154 V* 9.64853415 x10^4 = +29.717 KJ
Positive dG0 indicate non spontaneous reaction
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