Use tabulated electrode potentials to calculate ∆g° rxn for each reaction at 25 °c in kj.
(a) pb2+(aq) + mg(s) ➝ pb(s) + mg2+(aq)
(b) br2(l) + 2 cl-(aq) ➝ 2 br-(aq) + cl2( g)
(c) mno2(s) + 4 h+(aq) + cu(s) ➝ mn2+(aq) + 2 h2o(l) + cu2+(aq)

(a) , (b) and (c)  

Explanation: (a) Oxidation half reaction for the given equation is:

The reduction half equation is:

where n is the number of moles of electrons transferred and F is faraday constant.

2 moles of electrons are transferred in the cell reaction which is also clear from both the half equations.

or

(b) Oxidation half reaction for the given equation is:

Reduction half equation is:

Now, we can calculate the same as we did for part a.

or

(c) Oxidation half reaction for the given equation is:

reduction half equation is:

or

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.