Calculate the ratio of [cu2+]/[cu+] in an aqueous solution at eh = + 0.80 volt.

The ratio of in an aqueous solution is

Explanation:

The substance having highest positive potential will always get reduced and will undergo reduction react.

Oxidation half reaction:  

Reduction half reaction:      ( × 2)

Net cell reaction:  

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the of the reaction, we use the equation:

To calculate the EMF of the cell, we use the Nernst equation, which is:

Concentration of pure solids and liquids are taken as 1 in the ratio

where,

= electrode potential of the cell = +0.80 V

= standard electrode potential of the cell = -0.21 V

n = number of electrons exchanged = 2

= Ratio of the concentration

Putting values in above equation, we get:

Hence, the ratio of in an aqueous solution is

more info ? ? i don't understand what the question is

explanation:

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.