The organism would no longer grow.
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Neptune's atmosphere is composed up of many gases. These gases include Hydrogen (80%), Helium (19%), and Methane<span> (1.5%). Some of the minor gases include trace amounts of </span>Hydrogen Deuteride<span> and </span>Ethane<span>.</span>
Degree celsius is not an official si unit of measure.
<span>The mass of ni2 that is produced in solution by passing a current of 67.0 A for a period of 12.0 h and the cell is assumed as 85.0 % efficient is 766 Grams Ni. The answer to the question that was given above is 766 Grams Ni.</span>
