Consider a voltaic cell where the anode half-reaction is zn(s) → zn2+(aq) + 2 e− and the cathode half-reaction is sn2+(aq) + 2 e– → sn(s). what is the concentration of sn2+ if zn2+ is 2.5 × 10−3 m and the cell emf is 0.660 v? the standard reduction potentials are given below zn+2(aq) + 2 e− sn2+(aq) + 2 e– →→ zn(s) sn(s) e∘red e∘red == −0.76 v −0.136 v consider a voltaic cell where the anode half-reaction is and the cathode half-reaction is . what is the concentration of if is and the cell emf is 0.660 ? the standard reduction potentials are given below
a. 7.6 × 10−3 m
b. 9.0 × 10−3 m
c. 1.9 × 10−4 m
d. 6.9 × 10−4 m
e. 3.3 × 10−2 m

the answer is d : ) have a good da

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

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explanation:

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