Chemistry, 16.08.2020 14:01 Anaaguayo7406

Which metal(s) can be oxidized with a Pb2+ solution but not with a Cd2+ solution? (Hint: The reactions are occurring under standard conditions.) A. Cr
B. Ni
C. Zn
D. Sn

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Answer from: Leigh8655

B. Ni

D. Sn

Explanation:

Electrode Potential is the potential difference set up between an element and a solution of its ion. It is a measure of the tendency of an element to form ions.

The electrode potentials vary from one metal ion or metal system to another and the value depends on:

concentration of ions in the solution

the temperature at which the measurement is made , and

the overall energy change.

When two half-cells are joined together through a salt bridge, the e.m.f (electromotive force) of the cell formed is the algebraic difference between the two electrode potentials.

However, the set up in which chemical energy is converted to electrical energy is known as an Electrochemical cell. It consists of two half cells ;

an oxidation half-cell reaction

a reduction half cell reaction.

From the information given:

the standard reduction potential for each metal under standard conditions in the electrochemical series is as follows :

$E^0 _{Pb} = -0.126 \ V$

$E^0 _{Cr} = - 0.74 \ V$

$E^0 _{Ni} =- 0.23 \ V$

$E^0 _{Zn} =- 0.76 \ V$

$E^0 _{Sn} = -0.13 \ V$

$E^0 _{Cd} = - 0.40 \ V$

We will realize that Ni and Sn have reduction values in between Pb and Cd.

Thus , Ni can be oxidized by Pb2+ solution but not with a Cd2+ solution

Answer from: Quest

I’m not sure what your trying to say here.. can you clarify again?

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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Which metal(s) can be oxidized with a Pb2+ solution but not with a Cd2+ solution? (Hint: The reactio...

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