Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
<u>Answer:</u> The chemical reaction is given below.
<u>Explanation:</u>
A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.
The reactions which occur in hydrogen-oxygen fuel cell are:
At cathode: 
At anode: 
Net reaction: 
Thus, the chemical reaction is given above.
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
The correct Lewis structure for Fluorine is A.
To determine the Keq, we need the chemical reaction in the system. In this case it would be:
CO + 2H2 = CH3OH
The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:
CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13
Therefore,
Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51