Question

knowing that h₂ is produced during reduction and o₂ is produced during oxidation, what is the e°cell for the electrolysis of 1.0 m lif (aq)?
2h₂o + 2e⁻ → h₂ + 2oh⁻ e° = −0.83 v
o₂ + 4h⁺ + 4e⁻ → 2h₂o e° = +1.23 v
-2.89 v
2.06 v
2.89 v
-2.06 v

Explanation:

Step1: Identify Oxidation and Reduction Half - Reactions

For the electrolysis of aqueous LiF, we have to consider the possible half - reactions. The reduction reaction (where \(H_2\) is produced) is \(2H_2O + 2e^-\to H_2+2OH^-\) with \(E^{\circ}_{red}=- 0.83\ V\). The oxidation reaction (where \(O_2\) is produced) is the reverse of \(O_2 + 4H^++4e^-\to2H_2O\). When we reverse an oxidation - reduction half - reaction, the sign of \(E^{\circ}\) changes. So the oxidation half - reaction is \(2H_2O\to O_2 + 4H^++4e^-\) with \(E^{\circ}_{ox}=- 1.23\ V\) (since we reversed the given reduction half - reaction for oxidation).

Step2: Balance the Electrons and Calculate \(E^{\circ}_{cell}\)

First, we need to balance the number of electrons in the two half - reactions. The reduction half - reaction: \(2H_2O + 2e^-\to H_2+2OH^-\) (2 electrons transferred). The oxidation half - reaction: \(2H_2O\to O_2 + 4H^++4e^-\) (4 electrons transferred). To balance the electrons, we multiply the reduction half - reaction by 2: \(4H_2O + 4e^-\to2H_2 + 4OH^-\) with \(E^{\circ}_{red}\) (for the multiplied reaction) still \(-0.83\ V\) (because the standard reduction potential is an intensive property and does not change with the stoichiometric coefficient). The oxidation half - reaction remains \(2H_2O\to O_2 + 4H^++4e^-\) with \(E^{\circ}_{ox}=-1.23\ V\).

Now, the formula for \(E^{\circ}_{cell}\) is \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction)\) or \(E^{\circ}_{cell}=E^{\circ}_{red}(cathode)+E^{\circ}_{ox}(anode)\).

Using the second formula: \(E^{\circ}_{cell}=E^{\circ}_{red}(reduction\ half - reaction)+E^{\circ}_{ox}(oxidation\ half - reaction)\).

The reduction half - reaction (after multiplying by 2) has \(E^{\circ}_{red}=-0.83\ V\) and the oxidation half - reaction has \(E^{\circ}_{ox}=-1.23\ V\)? Wait, no. Wait, the correct formula is \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)+E^{\circ}_{anode}(oxidation)\). The cathode is where reduction occurs, and the anode is where oxidation occurs.

The reduction half - reaction (cathode): \(2H_2O + 2e^-\to H_2+2OH^-\), \(E^{\circ}_{cathode}=-0.83\ V\)

The oxidation half - reaction (anode): reverse of \(O_2 + 4H^++4e^-\to2H_2O\) is \(2H_2O\to O_2 + 4H^++4e^-\), \(E^{\circ}_{anode\ (oxidation)}=- 1.23\ V\) (since the original \(E^{\circ}\) for the reduction of \(O_2\) is \(+1.23\ V\), so for oxidation, it is \(-1.23\ V\))

But we need to balance the electrons. Let's multiply the reduction half - reaction by 2 to get 4 electrons: \(4H_2O + 4e^-\to2H_2+4OH^-\), \(E^{\circ}_{cathode}=-0.83\ V\) (intensive property, so no change with multiplication)

The oxidation half - reaction: \(2H_2O\to O_2 + 4H^++4e^-\), \(E^{\circ}_{anode\ (oxidation)}=-1.23\ V\)

Now, \(E^{\circ}_{cell}=E^{\circ}_{cathode}+E^{\circ}_{anode\ (oxidation)}\)

\(E^{\circ}_{cell}=(-0.83\ V)\times2+(- 1.23\ V)\)? No, that's wrong. Wait, the correct way is:

For a cell reaction, \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction)\)

The cathode reaction (reduction): \(2H_2O + 2e^-\to H_2+2OH^-\), \(E^{\circ}_{cathode}=-0.83\ V\)

The anode reaction (oxidation) is the reverse of \(O_2 + 4H^++4e^-\to2H_2O\), so the anode reaction (reduction) would be \(O_2 + 4H^++4e^-\to2H_2O\) with \(E^{\circ}_{anode}(reduction)=+1.23\ V\)

Then \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction)\)

\(E^{\circ}_{cell}=-0.83\ V-( + 1.23\ V)=-2.06\ V\)

Wait, let's re - do it. The reduction half - reaction (cathode): \(2H_2O + 2e^-\to H_2+2OH^-\), \(E^{\circ}_{red,cathode}=-0.83\ V\)

The oxidation half - reaction (anode): \(2H_2O\to O_2 + 4H^++4e^-\), \(E^{\circ}_{ox,anode}=-E^{\circ}_{red,anode}\), where \(E^{\circ}_{red,anode}\) is the reduction potential of the anode reaction. The anode reaction (reduction) is \(O_2 + 4H^++4e^-\to2H_2O\), \(E^{\circ}_{red,anode}=+1.23\ V\), so \(E^{\circ}_{ox,anode}=-1.23\ V\)

Now, to balance the electrons, we need to have the same number of electrons in both half - reactions. The reduction half - reaction has 2 electrons, the oxidation half - reaction has 4 electrons. Multiply the reduction half - reaction by 2: \(4H_2O + 4e^-\to2H_2+4OH^-\), \(E^{\circ}_{red,cathode}=-0.83\ V\) (since \(E^{\circ}\) is intensive)

The oxidation half - reaction: \(2H_2O\to O_2 + 4H^++4e^-\), \(E^{\circ}_{ox,anode}=-1.23\ V\)

Now, add the two half - reactions:

\(4H_2O + 4e^-+2H_2O\to2H_2 + 4OH^-+O_2 + 4H^++4e^-\)

Simplify: \(6H_2O\to2H_2 + O_2 + 4H^++4OH^-\), and \(4H^++4OH^-\to4H_2O\), so overall \(2H_2O\to2H_2 + O_2\)

Now, calculate \(E^{\circ}_{cell}\):

\(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)+E^{\circ}_{ox,anode}\)

The reduction half - reaction (after multiplying by 2) has \(E^{\circ}_{red}=-0.83\ V\) (per 2 electrons). The oxidation half - reaction has \(E^{\circ}_{ox}=-1.23\ V\) (per 4 electrons). Wait, no, the correct formula is \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction)\)

The cathode is where reduction occurs: \(2H_2O + 2e^-\to H_2+2OH^-\), \(E^{\circ}_{cathode}=-0.83\ V\)

The anode is where oxidation occurs, so the reduction reaction at the anode (reverse of oxidation) is \(O_2 + 4H^++4e^-\to2H_2O\), \(E^{\circ}_{anode}=+1.23\ V\)

\(E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=-0.83\ V - 1.23\ V=-2.06\ V\)

Answer:

\(-2.06\ V\) (corresponding to the option - 2.06 V)