The cell reaction 2Ag^+(aq)+H_2(g) rightarrow 2H^+(aq)+2Ag(s) has E degree_cell=

Question

The cell reaction 2Ag^+(aq)+H_2(g) rightarrow 2H^+(aq)+2Ag(s) has E degree_cell=0.80 V Under standard-state conditions, what is E degree for the following half-reaction? Ag^+(aQ)+e^- rightarrow Ag(s) Given: What is the standard Gibbs free-energy change (delta G degree) for the following reaction? For the cell reaction- the standard cell potential is 1.34 V. To determine the cell potential at nonstandard conditions, what is the value that should be used for n in the Nernst equation? How many faradays are required to convert one mole of S_2O_8^2- ions to SO_3^2- ions?

Explanation / Answer

There are two half reactions. they are H2(g) --------> 2H+ +2e- Eo1=0 (1)

Ag+ + e- -->Ag (2)

Multiply the Eq.2 with 2 to get 2Ag+ 2e- -------> 2Ag   Eo2=

When added, we geat

H2(g) + 2Ag+-------2Ag + 2H+ Eo= 0.8 = Eo1+Eo2

E02 =0.8 V ( A is correct)

b)

Given reactions are Cr+3 (aq) +3e- ------à Cr(s) E0= -.74V (1)

Multiplying Eq.1 with 2 and reversing it

2Cr --à 2Cr+3 + 6e-     Eo= .74V   (1A)

Pb+2 (aq)+ 2e- --------à Pb    Eo=-0.13V   (2)

Multipyling Eq. 2 with 3 gives, 3Pb+2 +6e- -----à 3Pb   Eo= -0.13V (2A)

Addition of Eq.1A and 2A gives

2Cr3+3Pb+2 ---------à 3Pb+ 2Cr+3 EO= 0.61V

delG= -nFE= -6*0.61*96500 J=-353190 Joules= -353 Kj

c) The half reactions can be written as

5e- + MnO4- + 8H+ ------->Mn+2 + 4H2O   (1)

H2O + H2SO3---------> SO4-2 +4H+ +2e- (2)

for balcing the reaction, Eq.1 needs to be multiplied with 2 and Eq. 2 with 5. This gives rise to 10 electrons. ( B is correct)

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