Question

5. Given the cell notation Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s), what is the half-reaction that occurs at the anode?

Cu2+(aq) + 2e–→ Cu(s)
Cu(s) → Cu2+(aq) + 2e–
Ag+(aq) + e–→ Ag(s)
Ag+(aq) + Cu(s) → Ag(s) + Cu2+(aq)


6. What is the E0 for the spontaneous reaction when a Ni2+/Ni half-cell is joined to a Cu2+/Cu half-cell? E0Cu= + 0.34 V = +0.34 V and E0Ni= -0.26 V.

–0.08 V
+0.08 V
+0.60 V
–0.60 V

Answer 1

What is [H+] given that the measured cell potential is -0.464 V and the anode reduction ... What half-reaction occurs at the cathode during the electrolysis of molten ... PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l); E° = 1.69 V .... For the cell Cu(s)|Cu2+||Ag+|Ag(s), the standard cell potential is 0.46 V. A cell ... hopw this helps

Answer 2

1) The half-reaction that occurs at the anode is :

`Cu(s)\rightarrow Cu^(2+)+2e^-`

2)+0.60 V is the E° for the spontaneous reaction.

Step-by-step explanation:

1)

Oxidation reaction is defined as the reaction in which an atom looses its electrons.

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

Oxidation reaction occurs at anode.

Oxidation half reaction:

`Cu(s)\rightarrow Cu^(2+)(aq)+2e^-`

Reduction reaction occurs at cathode.

Reduction half reaction:

`Ag^(+)(aq)+e^-\rightarrow Ag(s)`

Net reaction:
`Cu(s)+2Ag^(+)(aq)\rightarrow Cu^(2+)(aq)+2Ag(s)`

2)

Reduction potential of nickel(II) ions to nickel =
`E^o_(Ni^(2+)/Ni)=-0.26 V`

Reduction potential of copper (II) ions to copper =
`E^o_(Cu^(2+)/Cu)=0.34 V`

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(red,cathode)-E^o_(red,anode)`

Substance with higher or positive reduction potential goes on cathode and substance with lower of more negative reduction potential goes on anode.

Putting values in above equation, we get:

`E^o_(cell)=0.34 V-(-0.26)=0.60 V`

+0.60 V is the E° for the spontaneous reaction.