Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
<span>Who can help me with a few chemistry balancing equations?
Me. Just show me the equations :)</span>
Answer:
1.
%K = 31.904%
%Cl = 28.930%
%O = 39.166%
2.
Hg = 80.69/ 200.59 = 0.40/0.2 = 2
S = 6.436/ 32.07 = 0.20/0.2 = 1
O = 12.87/ 16.00 = 0.80/0.2 = 4
Hg2SO4 is the empirical formula
Explanation:
brainliest please?
All of the group iia metals (the column with Be, Mg, Ca, etc) have an outermost s suborbital with 2 electrons
When an atoms oxidation number decreases in a reaction then it’s being reduced. the answer for that question is B
