When a magnesium wire is dipped into a solution of lead (II) nitrate, a black deposit forms on the wire. Which of the following
can be concluded from this observation?
(A) The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
(B) Mg(s) is less easily oxidized than Pb(s).
(C) An external source of potential must have been supplied.
(D) The magnesium wire will be the cathode of a Mg/Pb cell.
(E) Pb(s) can spontaneously displace Mg2+(aq) from solution.
...
(A) The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
(B) Mg(s) is less easily oxidized than Pb(s).
(C) An external source of potential must have been supplied.
(D) The magnesium wire will be the cathode of a Mg/Pb cell.
(E) Pb(s) can spontaneously displace Mg2+(aq) from solution.
...
1 answer:
Answer:The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
Explanation:
Metals are usually arranged in an order of reactivity called activity series. Metals that are high up in the series are good reducing agents with very low (very negative) reduction potentials. Metals with greater (less negative) reduction potentials are found lower in the series. In the image attached, elements were arranged according to their reducing ability. Magnesium is very electro positive hence it is a better reducing agent with a lesser standard reduction potential than lead(refer to the image for numerical values of standard reduction potentials). Hence it displaces lead from solution and the elemental lead deposits on the wire.
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Answer:
Explanation:
From the given information;
The chemical reaction can be well presented as follows:
⇄
Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:
i.e
However, since we are dealing with liquids solutions;
since the activity of
is equivalent to 1
Hence, under standard conditions(i.e at a pressure of 1 bar)
(b)
From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:
Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol
(c)
Le's recall that:
At equilibrium, the instantaneous free energy is usually zero &
Q(reaction quotient) is equivalent to K(equilibrium constant)
So;
(d)
The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).
This is because;
If Q < K, then the reaction will proceed in the right direction towards the products.
However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.
So;
Since we are dealing with liquids;
Q = 1
Since Q < K; Then, the reaction proceeds in the right direction.
Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.
Answer:
Explanation:
We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.
M_r: 58.32
Mg(OH)₂ + … ⟶ … + 2HOH
m/g: 58.3
(a) Moles of Mg(OH)₂
(b) Moles of H₂O
The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.
The reaction will form of water.