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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the answer would be a