Consider a galvanic cell based on the reaction al^3+_(aq) + mg_(s) rightarrow al_(s) + mg^2+ _(aq) the half-reactions are al^3+ + 3 e^- rightarrow al e degree = - 1.66 v mg^2+ + 2 e^- rightarrow mg e degree = - 2.37 v give the balanced cell reaction and calculate e degree for the cell.

The standard cell potential of the cell is -0.71 V

Explanation:

The half reactions follows:

Oxidation half reaction:    ( × 3)

Reduction half reaction:    ( × 2)

The balanced cell reaction follows:

To calculate the of the reaction, we use the equation:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

Hence, the standard cell potential of the cell is -0.71 V

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).