The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn2+ ions. These ions are then oxidized to the deeply colored MnO4- ions by periodate ion (IO4-) in acid solution. Calculate ℰ° and ΔG° at 25°C for each reaction

The equations are:

3Mn + 2NO3- + 8H+ > 3Mn2+ + 2NO + 2H2O

2Mn2+ + 5IO4- + 3H2O > 2MnO4- + 5IO3-+ 6H

Ecell = 1.6 V - 1.51 V = 0.09 V

Using the equation of Nernst ,

 ΔG = -nFEcell

where n= number of electrons transferred = 10 mol

F = Farraday's constant = 96485 C/mol

Ecell = 0.09V

Now,   ΔG = -10 * 96485 C/mol * 0.09 V

ΔG = -86836.5 J/mol (\because C*V =J)

ΔG = -86.84 kJ/mol

Now, Ecell = (0.0591/n)*log K

0.09 = 0.0591/10 *logK

0.09 = 0.00591 * log K

log K = 0.09 / 0.00591

log K = 15.23

K = 10-15.23

  K = 5.89*10-16

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

it should be all of them because it asks if you included all of those in your answer

explanation: