Question

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Co^{2+}(aq) + H_{2}(g) ? Co(s) + 2H^{+}(aq) Co^{2+}(aq) + 2e^{-} ? Co(s) | E^{o}_{red} = -0.280 V 2H^{+}(aq) + 2e^{-} ? H_{2}(g) | E^{o}_{red} = 0.000 V ?G^{o} = kJ K for this reaction would be than one.

          Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
Co^{2+}(aq) + H_{2}(g) ? Co(s) + 2H^{+}(aq)
Co^{2+}(aq) + 2e^{-} ? Co(s) | E^{o}_{red} = -0.280 V
2H^{+}(aq) + 2e^{-} ? H_{2}(g) | E^{o}_{red} = 0.000 V
?G^{o} = kJ
K for this reaction would be than one.

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Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
Co^2+(aq) + H2(g) ? Co(s) + 2H^+(aq)
Co^2+(aq) + 2e^- ? Co(s) | E^ored = -0.280 V
2H^+(aq) + 2e^- ? H2(g) | E^ored = 0.000 V
?G^o = kJ
K for this reaction would be than one.

Added by Makeda G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Dominique Jan Tan

11/23/2022

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Transcript

00:01 The standard free energy change of an electrochemical reaction can be computed using the formula delta g not equals negative nf where e knot is the cell potent is the standard cell potential and is the number of electrons transferred and f is the paradise constant so our first the first thing that we do here is to find the standard cell potential from the given half reactions so this will solve us the different between the reduction potential at the cathode and the reduction potential at the anode with the reduction potential of the cathode as the one that is more positive than the reduction potential at the cathode so looking at the given values the more positive and that's the one that is the reduction potential at the cathode is the second half reaction with a reduction potential of zero and the other reaction with a potential of negative or the reduction potential of negative point 28 will be the reduction potential for their anodic reaction so solving this use as 0 .280 volts and now let's looking at the reaction here so we find that two electrons are are involved so that means two electrons will be transferred and so your number of n will be equal to 2.

01:34 So again this is the number of electrons transferred and we find that from the reaction from the half reactions.

01:42 So number of electron.

01:47 So now we can solve for delta g that is equal to negative n which is 2.

01:53 F which is 96 ,500.

01:56 This is the charge of one moon of electron and this is called your finalities constant multiplied by your cell potential standard cell potential so this gives us the standard free energy change of a reaction in the unit of juice juice per mole or just juice is as equal to negative 5 ,000 or 54 ,040 and then we can convert this in kilo juice by dividing it by 1 ,000 since one kilojoules is a thousand juice and this gives us delta gene nut of negative 54 .0.

02:41 So this is the delta g nut.

02:44 And with the knowledge of delta g nut, we can also solve for the equilibrium constant.

02:50 Recalling that delta genot is equal to negative r t and n of k, where r is the gas constant and t is the temperature...

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