AGo and Eo can be said to measure the same thing, and are convertible by the equ

Question

AGo and Eo can be said to measure the same thing, and are convertible by the equation AG… _nFE cell where n is the total number of moles of electrons being transferred, and Fis the Faraday constant 9.64853415×104 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 Mn (II) and the anode is Sc(III) free energy spontaneity Select kJ (b) The cathode is Zn(II) and the anode is Sn(II) free energy 402 spontaneity Select- k) (c) The cathode is Pb(II) and the anode is Cd() free energy 49 spontaneity Select kT

Explanation / Answer

The E call is calculated by the difference of the E cathode - E anode . In the cathode the reduction takes place , and the oxidation is done at the anode.

The gibbs free energy of the reeaction will be determined by the sign of the E call , the positive value will make the overall G negative making it spontaneous and vice-versa.

Mn (2+) = -1.185 and the

0.892V.

the next E call would be calculated similarly. Zn(2+) to Zn reduction potential is

and the Sn(2+) is -1.07. The E cell is -0.7618-(-1.07) = 0.3082 V.

The third example is Pb(2+) is -0.58 and the Cd(2+) is -0.40 V.

The E celll is calculated same way as the E cathode - E anode = -0.58+0.40 V = -0.18 V.

The spontaneity or the Gibbs free energy is dictated by the sign of the E cell , the negative E cell makes the reaction Gibbs free energy positive(non-spontaneous) and the positive G is decided by the negative E cell.

So the first and the second answers are non-spontaneous and the third reaction is spontaneous.

The value of Free energy is calculated by adding nF term. For the first question the n= 6, and multiplynig by F=96485 gives 579Kjoules. Now again multiplying by the E cell give the value 516.462 kJ.

The second and third questions have n = 2 both. The nF term gives 193 kJ and multiplying by the respective E cell gives the values 589.48 and -34.74.

Plz note that the last reaction is spontaneous (due to the negative sign) and the free energy values are approximated.

Sc3+ + 3e Sc(s) E = 2.077 V. The E cell = -1.185- (-2.077) =

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