Chemistry, 28.01.2020 15:53 idk5233

Which of these redox reactions is spontaneous as written?

ni + zn2+ → ni2+ + zn

mg + li+ → mg2+ + li

cu + au3+ → cu2+ + au

zn + ca2+ → zn2+ + ca

Which of these redox reactions is spontaneous as written? ni + zn2+ → ni2+ + zn

Answers: 3

Show answers

Answers

Answer from: RidhaH

$Cu+Au^{3+}\rightarrow Cu^{2+}+Au$

Explanation: when E^0 = +ve, reaction is spontaneous

E^0 = -ve, reaction is non spontaneous

E^0 = 0, reaction is in equilibrium

1. $Ni+Zn^{2+}\rightarrow Ni^{2+}+Zn$

Here Ni undergoes oxidation by loss of electrons, thus act as anode. zinc undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both E^0 are standard reduction potentials.

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0_{[Ni^{2+}/Ni]}= -0.26V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Ni^{2+}/Ni]}$

E^0=-0.76V- (-0.26V)=-0.50V

As the standard emf comes to be negative, the reaction is non spontaneous.

2) $Mg+Li^{+}\rightarrow Mg^{2+}+Li$

Here Mg undergoes oxidation by loss of electrons, thus act as anode. Lithium undergoes reduction by gain of electrons and thus act as cathode.

$E^0_{[Mg^{2+}/Mg]}= -2.37V$

$E^0_{[Li^{+}/Li]}= -3.05V$

$E^0=E^0_{[Li^{+}/Li]}- E^0_{[Mg^{2+}/Mg]}$

E^0=-3.05- (-2.37V)=-0.68V

As the standard emf comes to be negative, the reaction is non spontaneous.

3) $Cu+Au^{3+}\rightarrow Cu^{2+}+Au$

Here Cu undergoes oxidation by loss of electrons, thus act as anode. Gold undergoes reduction by gain of electrons and thus act as cathode.

$E^0_{[Au^{3+}/Au]}= +1.40V$

$E^0_{[Cu^{2+}/Cu]}= 0.34V$

$E^0=E^0_{[Au^{3+}/Au]}- E^0_{[Cu^{2+}/Cu]}$

E^0=1.40- (+0.34V)=1.06V

As the standard emf comes to be positive, the reaction is spontaneous.

4) $Zn+Ca^{2+}\rightarrow Zn^{2+}+Ca$

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Calcium undergoes reduction by gain of electrons and thus act as cathode.

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0_{[Ca^{2+}/Ca]}=-2.87V$

$E^0=E^0_{[Ca^{2+}/Ca}- E^0_{[Zn^{2+}/Zn]}$

E^0=-2.87V- (-0.76V)=-2.11V

As the standard emf comes to be negative, the reaction is non spontaneous.

Answer from: Quest

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:

Answer from: Quest

yes

ya te ayud espera