First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
The inputs of photosynthesis are light energy, and matter in the form of water absorbed through the roots, and carbon dioxide absorbed through the leaves.
Therefore an input of photosynthesis from the choices given is,
Sunlight
0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.
Answer:
b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.
d. It should be disposed of in the bottle for waste copper ion when work is completed.
Explanation:
Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.
It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.
In the Solutions and Spectroscopy experiments there is always wastes.
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
