Calculate the standard free-energy change for the reaction at 25 ∘ c. 25 ∘c. refer to the list of standard reduction potentials. 2 au 3 + (aq) + 3 ni (s) − ⇀ ↽ − 2 au (s) + 3 ni 2 + (aq) 2au3+(aq)+3ni(s)↽−−⇀2au(s)+3ni2+(aq ) δ g ∘ = δg∘=

The for the given reaction is

Explanation:

For the given chemical reaction:

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.

Oxidation half reaction:     ( × 3)

Reduction half reaction:     ( × 2)

We know that:

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:

Relationship between standard Gibbs free energy and standard electrode potential follows:

where,

n = number of electrons transferred = 6

F = Faraday's constant = 96500 C

= standard cell potential = 1.73 V

Putting values in above equation, we get:

Hence, the for the given reaction is

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year