A student made measurements on some electrochemical cells and calculated three q

Question

A student made measurements on some electrochemical cells and calculated three quantities The standard reaction free energy G . The equilibrium constant K at 25.0 oC The cell potential under standard conditions E His results are listed below Unfortunately, the student may have made some mistakes. Examine his results carefully and tick the box next to the incorrect quantity in each row, if any Note: If there is a mistake in a row, only one of the three quantities listed is wrong. Also, you may assume the number of significant digits in each quantity is correct Also note: for each cell, the number n of electrons transferred per redox reaction is 1 calculated quantities (Check the box next to any that are wrong.) cell n 63 kJ/mol Ol 129. kJ/mol O 59. kJ/mol| 1.09 × 10 3.98 × 10 2.17×10 -0.65 V 1.34 V 10 O -0.61 V

Explanation / Answer

for the first case delta G0 = 63 kj

K = 1.09 * 10^11

so as we know the relation of delta G0 with K

Delta G0 = - RT lnK

= - (8.314 J mol-1 K-1) (298 K) ln (1.09 *10^11)

= - 62966.5 J

= -63 KJ / mol (approximate)

Agin Delta G0 = -nFE0

G0= -1* 96500*(-0.65)      ;   F= Faradeys constant

G0 = 62.725 KJ

There is wrong in value of K = 1.09 *10^ -11

B:

Delta G0 = - RT lnK

= - (8.314 J mol-1 K-1) (298 K) ln (3.98 *10^22)

= - 128924.5 J

= -129 KJ / mol (approximate)

Agin Delta G0 = -nFE0

G0 = -1* 96500*1.34      ;   F= Faradeys constant

G0 = -129.3 KJ

Thre is wrong in sign in G0

C:

Delta G0 = - RT lnK

= - (8.314 J mol-1 K-1) (298 K) ln (2.17 *10^10)

= - 58965.9 J

= -59. KJ / mol (approximate)

Agin Delta G0 = -nFE0

G0 = -1* 96500*-0.61      ;   F= Faradeys constant

G0 = 59 KJ

There was wrong data in G0 it could be -59KJ or

Wrong data in K = 2.17*10^-10

= -

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