Edexcel IAL Chemistry A2 Topic 16 Redox Equilibria

[Maximum number: 9]

This question is about mercury, Hg, and its compounds.
Mercury is a liquid element in the same group of the Periodic Table as zinc.
The electronic configuration of mercury is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^{2}$ .

(a)

Mercury reacts with nitric acid to form an aqueous solution of $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ and nitrogen monoxide gas.

The unbalanced equation is shown.

\mathrm{Hg}(\mathrm{l})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NO}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

[ 2 ]

(i)

Explain, using oxidation numbers, why this is a redox reaction.
(2)

[ 2 ]

An explanation that makes reference to the following
points:

Allow oxidation numbers from annotation to the equation
Ignore any reference to electron loss/gain

Expert

- Hg/mercury oxidised and
from 0 (in Hg ) to +2 (in $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ )

(1)

Do not award reference to oxidation of any other element

- N / nitrogen is reduced
and
from +5 (in $\mathrm{HNO}_{3}$ ) to +2 (in NO)

(1)

Do not award $\mathrm{HNO}_{3}$ is reduced
Do not award reference to reduction of any other element

If no other mark awarded,
Hg/mercury oxidised and N/nitrogen reduced
OR
Hg / mercury from 0 to +2 and N / nitrogen from +5 to +2
scores (1)

(b)

Mercury(I) chloride, $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$ , is also known as calomel.

A saturated calomel electrode may be used as an alternative to the standard hydrogen electrode.

Figure for Question (b) — Edexcel A-Level Chemistry A2

The half-equation for the calomel electrode is

\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Hg}(\mathrm{l})+2 \mathrm{Cl}^{-}(\mathrm{aq})

The standard electrode potential of the calomel electrode is $E^{\ominus}=+0.24 \mathrm{~V}$ .

[ 7 ]

(i)

Suggest why KCl crystals are needed in the outer tube of the electrode.

[ 1 ]

An answer that makes reference to the following point:

- (to provide a) constant concentration (of $\mathrm{Cl}^{-}$ )

Allow KCl for $\mathrm{Cl}^{-}$ throughout

Allow to keep the solution / $\mathrm{Cl}^{-}$ saturated

Allow to replace $\mathrm{Cl}^{-}$

Ignore just to provide $\mathrm{Cl}^{-}$ Ignore stated concentrations

Do not award salt bridge / to complete the circuit

Do not award catalyst

(1)

Expert

(ii)

A calomel electrode was used to measure the standard electrode potential of the $\mathrm{Sn}^{2+}(\mathrm{aq}) \mid \mathrm{Sn}(\mathrm{s})$ half-cell.

Figure for Question (ii) — Edexcel A-Level Chemistry A2

The reading on the voltmeter in this cell was +0.37 V .
Deduce the standard electrode potential for the $S n^{2+}(a q) \mid S n(s)$ half-cell.

[ 1 ]

An answer that makes reference to the following point:

- $(0.24-0.37=)-0.13(\mathrm{~V})$

Ignore working, even if incorrect

Expert

(iii)

Write the overall equation for the cell reaction.

[ 1 ]

An answer that makes reference to the following point:
$\text { - } \mathrm{Hg}_{2} \mathrm{Cl}_{2}+\mathrm{Sn} \rightarrow 2 \mathrm{Hg}+\mathrm{Sn}^{2+}+2 \mathrm{Cl}^{-}$

Allow $\mathrm{Hg}_{2} \mathrm{Cl}_{2}+\mathrm{Sn} \rightarrow 2 \mathrm{Hg}+\mathrm{SnCl}_{2}$ Allow multiples

Allow ⇌ for →

Ignore state symbols even if incorrect Ignore half-equations even if incorrect Ignore use of cell diagrams

Do not award uncancelled electrons

Do not award $2 \mathrm{Hg}^{+}\left(+2 \mathrm{Cl}^{-}\right)$ for $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$

If answer to (d)(ii) is $+0.61(\mathrm{~V}) /+0.37(\mathrm{~V}) /$ greater than $+0.24(\mathrm{~V})$ , equation must be reversed:

$2 \mathrm{Hg}+\mathrm{Sn}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}+\mathrm{Sn}$ OR

$2 \mathrm{Hg}+\mathrm{SnCl}_{2} \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}+\mathrm{Sn}$

Expert

(iv)

Add labels to complete the diagram of a standard hydrogen electrode. Include details of any essential conditions.
(3)

Figure for Question (iv) — Edexcel A-Level Chemistry A2

[ 3 ]

(v)

Suggest one advantage of using a calomel electrode, in place of a standard hydrogen electrode, when measuring a cell potential.

[ 1 ]

[Maximum number: 5]

An electrochemical cell is made from the electrode systems shown by these half-equations.

\begin{aligned} \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{Ag}(\mathrm{~s}) & E^{\ominus}=+0.80 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & E^{\ominus}=+1.00 \mathrm{~V} \end{aligned}

The apparatus used to measure the value for $E_{\text {cell }}^{\ominus}$ under standard conditions is shown.

Figure for Question 2 — Edexcel A-Level Chemistry A2

(a)

Which silver compound could be used as the electrolyte in the left-hand half-cell?

A

silver nitrate

B

silver hydroxide

C

silver chloride

D

silver carbonate

[ 1 ]

(b)

The electrolyte in the right-hand half-cell is prepared using equal volumes of

A

$1 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}^{2+}(\mathrm{aq})$

B

$2 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $2 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}^{2+}(\mathrm{aq})$

C

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}(\mathrm{aq})$

D

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{VO}^{2+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}(\mathrm{aq})$

[ 1 ]

(c)

Which is the equation for the overall cell reaction under standard conditions?

A

$\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{VO}_{2}^{+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})$

B

$\mathrm{VO}_{2}^{+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

C

$\mathrm{VO}^{2+}(\mathrm{aq})+3 \mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{VO}_{2}^{+}(\mathrm{aq})+3 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

D

$\mathrm{VO}_{2}^{+}(\mathrm{aq})+3 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{VO}^{2+}(\mathrm{aq})+3 \mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})$

[ 1 ]

(d)

Which is the value of $E_{\text {cell }}^{\ominus}$ in volts?

A

-1.80

B

-0.20

C

+0.20

D

+1.80

[ 1 ]

(e)

Which is the cell diagram for this cell, using the conventional representation of half-cells?

A

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right]\right|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right] \mid \mathrm{Pt}(\mathrm{s})$

B

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right],\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]\right| \mathrm{Pt}(\mathrm{s})$

C

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]\right|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right] \mid \mathrm{Pt}(\mathrm{s})$

D

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right],\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right]\right| \mathrm{Pt}(\mathrm{s})$

[ 1 ]

[Maximum number: 12]

This question is about cell reactions involving chromium.
Use your Data Booklet when answering this question.

(a)

Name the type of the forward reaction that is shown in the right-hand electrode systems in the Data Booklet.

[ 1 ]

(b)

Name the series formed when the right-hand electrode systems are placed in order, most negative first.

[ 1 ]

(c)

A student set up the following apparatus to measure the standard electrode potential for right-hand electrode system 8 in your Data Booklet.

Figure for Question (c) — Edexcel A-Level Chemistry A2

[ 6 ]

(i)

Identify X and Y.

X:

Y:

[ 2 ]

platinum / Pt

- $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{2+}$ and

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{3+}$

Accept platinised platinum / platinum black

Allow $1 \mathrm{~mol} \mathrm{dm}^{-3}$ named soluble chromium(II) salts

Allow named soluble chromium(III) salts giving $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{3+}$ solution

Allow $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{2+}$ and $\mathrm{Cr}^{3+}$

Ignore equimolar

(ii)

Give two reasons why the initial reading on the voltmeter may differ from the stated value given in the Data Booklet.

[ 2 ]

(iii)

The voltmeter is removed and the cell is allowed to run for one hour.

Explain the changes that would occur in the right-hand half-cell during this time.

[ 2 ]

An explanation that makes reference to the following points:

- the (concentration of) $\mathrm{Cr}^{2+}$ decreases / $\mathrm{Cr}^{3+}$ increases (1)

- because $E_{\text {cell }}$ is positive $/+0.41 \mathrm{~V}$

(1)

Allow the blue-green colour becomes green(er)

Allow correct half equation of oxidation

Allow $\mathrm{Cr}^{2+}$ oxidises/changes to $\mathrm{Cr}^{3+}$

Ignore colours if concentration is discussed

Ignore equilibrium shifts to the left alone

Ignore comments relating to salt bridge

(d)

Explain, by calculating $E_{\text {cell }}^{\ominus}$ values, why Fe (II) is used to reduce Cr (VI) to Cr (III) but zinc is used to reduce $\operatorname{Cr}(\mathrm{VI})$ to Cr(II).
(4)

Numbers chosen of the right-hand electrode systems from the Data Booklet
Explanation

[ 4 ]

Answer that includes the following:

(5, 8, 21, 28)

- zinc converts dichromate(VI) ions to chromium(III) ions with an $E_{\text {cell }}^{\ominus}$ value of +2.09 V

(1)

- and then to chromium(II) ions with an $E_{\text {cell }}^{\ominus}$ value of $+0.35 \mathrm{~V}$

(1)

- iron(II) converts dichromate(VI) ions to chromium(III) ions with an $E_{\text {cell }}^{\ominus}$ value of +0.56 V

(1)

- but not to chromium(II) ions as $E_{\text {cell }}^{\ominus}$ value is -1.18 V which is negative

(1)

If no other mark is awarded allow one mark for the four numbers in any order 5,8,21,28 or values -0.76 V, $-0.41 \mathrm{~V},+0.77 \mathrm{~V}$ and +1.33 V

Accept equations in place of descriptions

Allow positive numbers without + sign

No TE on incorrect equations being used

[Maximum number: 2]

This question is about mercury, Hg, and its compounds.
Mercury is a liquid element in the same group of the Periodic Table as zinc.
The electronic configuration of mercury is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^{2}$ .

Mercury reacts with nitric acid to form an aqueous solution of $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ and nitrogen monoxide gas.

The unbalanced equation is shown.

\mathrm{Hg}(\mathrm{l})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NO}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Explain, using oxidation numbers, why this is a redox reaction.
(2)

An explanation that makes reference to the following
points:

Allow oxidation numbers from annotation to the equation
Ignore any reference to electron loss/gain

Expert

- Hg/mercury oxidised and
from 0 (in Hg ) to +2 (in $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ )

(1)

Do not award reference to oxidation of any other element

- N / nitrogen is reduced
and
from +5 (in $\mathrm{HNO}_{3}$ ) to +2 (in NO)

(1)

Do not award $\mathrm{HNO}_{3}$ is reduced
Do not award reference to reduction of any other element

If no other mark awarded,
Hg/mercury oxidised and N/nitrogen reduced
OR
Hg / mercury from 0 to +2 and N / nitrogen from +5 to +2
scores (1)

[Maximum number: 1]

An electrochemical cell is made from the electrode systems shown by these half-equations.

\begin{aligned} \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{Ag}(\mathrm{~s}) & E^{\ominus}=+0.80 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & E^{\ominus}=+1.00 \mathrm{~V} \end{aligned}

The apparatus used to measure the value for $E_{\text {cell }}^{\ominus}$ under standard conditions is shown.

Figure for Question 5 — Edexcel A-Level Chemistry A2

The electrolyte in the right-hand half-cell is prepared using equal volumes of

A

$1 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}^{2+}(\mathrm{aq})$

B

$2 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $2 \mathrm{~mol} \mathrm{dm}^{-3}$ acidified $\mathrm{VO}^{2+}(\mathrm{aq})$

C

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{VO}_{2}^{+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}(\mathrm{aq})$

D

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{VO}^{2+}(\mathrm{aq})$ and $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}(\mathrm{aq})$

[Maximum number: 2]

This question is about cell reactions involving chromium.
Use your Data Booklet when answering this question.

A student set up the following apparatus to measure the standard electrode potential for right-hand electrode system 8 in your Data Booklet.

Figure for Question 6 — Edexcel A-Level Chemistry A2

Identify X and Y.

X:

Y:

platinum / Pt

- $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{2+}$ and

$1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{3+}$

Accept platinised platinum / platinum black

Allow $1 \mathrm{~mol} \mathrm{dm}^{-3}$ named soluble chromium(II) salts

Allow named soluble chromium(III) salts giving $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{3+}$ solution

Allow $1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Cr}^{2+}$ and $\mathrm{Cr}^{3+}$

Ignore equimolar

[Maximum number: 3]

This question is about copper and some of its compounds.

An electrochemical cell was made from the electrode systems represented by these half-equations:

\begin{array}{ll} \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Cu}(\mathrm{~s}) & E^{\ominus}=+0.34 \mathrm{~V} \\ \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}^{2+}(\mathrm{aq}) & E^{\ominus}=+0.77 \mathrm{~V} \end{array}

A student drew a diagram of an experiment to measure the standard emf of the cell.

Figure for Question 7 — Edexcel A-Level Chemistry A2

Identify three mistakes in this diagram and the changes needed to correct them.

Assume that standard conditions were used.

Table for Question 7 — Edexcel A-Level Chemistry A2

[Maximum number: 1]

This question is about mercury, Hg, and its compounds.
Mercury is a liquid element in the same group of the Periodic Table as zinc.
The electronic configuration of mercury is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^{2}$ .

Mercury(I) chloride, $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$ , is also known as calomel.

A saturated calomel electrode may be used as an alternative to the standard hydrogen electrode.

Figure for Question 8 — Edexcel A-Level Chemistry A2

The half-equation for the calomel electrode is

\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{Hg}(\mathrm{l})+2 \mathrm{Cl}^{-}(\mathrm{aq})

The standard electrode potential of the calomel electrode is $E^{\ominus}=+0.24 \mathrm{~V}$ .

A calomel electrode was used to measure the standard electrode potential of the $\mathrm{Sn}^{2+}(\mathrm{aq}) \mid \mathrm{Sn}(\mathrm{s})$ half-cell.

The reading on the voltmeter in this cell was +0.37 V .
Deduce the standard electrode potential for the $S n^{2+}(a q) \mid S n(s)$ half-cell.

An answer that makes reference to the following point:

- $(0.24-0.37=)-0.13(\mathrm{~V})$

Ignore working, even if incorrect

Expert

[Maximum number: 1]

An electrochemical cell is made from the electrode systems shown by these half-equations.

\begin{aligned} \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{Ag}(\mathrm{~s}) & E^{\ominus}=+0.80 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} & \rightleftharpoons \mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & E^{\ominus}=+1.00 \mathrm{~V} \end{aligned}

The apparatus used to measure the value for $E_{\text {cell }}^{\ominus}$ under standard conditions is shown.

Figure for Question 9 — Edexcel A-Level Chemistry A2

Which is the cell diagram for this cell, using the conventional representation of half-cells?

A

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right]\right|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right] \mid \mathrm{Pt}(\mathrm{s})$

B

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right],\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]\right| \mathrm{Pt}(\mathrm{s})$

C

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]\right|\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right] \mid \mathrm{Pt}(\mathrm{s})$

D

$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}(\mathrm{aq})\right|\left|\left[\mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right],\left[\mathrm{VO}_{2}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})\right]\right| \mathrm{Pt}(\mathrm{s})$

[Maximum number: 2]

This question is about cell reactions involving chromium.
Use your Data Booklet when answering this question.

A student set up the following apparatus to measure the standard electrode potential for right-hand electrode system 8 in your Data Booklet.

Figure for Question 10 — Edexcel A-Level Chemistry A2

Give two reasons why the initial reading on the voltmeter may differ from the stated value given in the Data Booklet.

[Maximum number: 2]

Brass is an alloy of copper. The percentage by mass of copper in a sample of brass can be determined by a three-stage process.

Stage 1 Oxidation of the copper to copper(II) ions with excess concentrated nitric acid.
Stage 2 Reduction of the copper(II) ions to copper(I) ions with excess iodide ions.
Stage 3 Titration of the iodine produced in Stage 2 against a standard solution of sodium thiosulfate.

In Stage 2, $25.0 \mathrm{~cm}^{3}$ portions of the solution from Stage 1 are transferred into conical flasks.
$10 \mathrm{~cm}^{3}$ of a solution of potassium iodide is added to each flask and the mixtures are swirled. Copper(I) iodide is precipitated.

The redox equation for the reaction between copper(II) ions and iodide ions is shown.

2 \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{aq})

The standard electrode potentials for the half-equations are
Reduction $\mathrm{Cu}^{2+}+\mathrm{e}^{-} \rightleftharpoons \mathrm{Cu}^{+} \quad E^{\ominus}=+0.15 \mathrm{~V}$
Oxidation $\mathrm{I}_{2}+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{I}^{-} \quad E^{\ominus}=+0.54 \mathrm{~V}$

Use the electrode potential data to explain why the redox reaction might not be expected to be feasible.

adding electrode potentials

- cell potential is negative /<0 (so not feasible)

Example of calculation
$E_{\text {cell }}=+0.15-0.54=-0.39(V)$

Incorrect negative Ecell value can score M2

[Maximum number: 1]

Which graph shows the relationship between Ecell plotted on the x-axis and ΔStotal on the y-axis for any electrochemical system?

Figure for Question 12 — Edexcel A-Level Chemistry A2

A

graph A

B

graph B

C

graph C

D

graph D

[Maximum number: 2]

Brass is an alloy of copper. The percentage by mass of copper in a sample of brass can be determined by a three-stage process.

Stage 1 Oxidation of the copper to copper(II) ions with excess concentrated nitric acid.
Stage 2 Reduction of the copper(II) ions to copper(I) ions with excess iodide ions.
Stage 3 Titration of the iodine produced in Stage 2 against a standard solution of sodium thiosulfate.

In Stage 2, $25.0 \mathrm{~cm}^{3}$ portions of the solution from Stage 1 are transferred into conical flasks.
$10 \mathrm{~cm}^{3}$ of a solution of potassium iodide is added to each flask and the mixtures are swirled. Copper(I) iodide is precipitated.

The redox equation for the reaction between copper(II) ions and iodide ions is shown.

2 \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{aq})

The standard electrode potentials for the half-equations are
Reduction $\mathrm{Cu}^{2+}+\mathrm{e}^{-} \rightleftharpoons \mathrm{Cu}^{+} \quad E^{\ominus}=+0.15 \mathrm{~V}$
Oxidation $\mathrm{I}_{2}+2 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{I}^{-} \quad E^{\ominus}=+0.54 \mathrm{~V}$

Explain why the redox reaction does, in fact, take place.

the copper (I) iodide/ $\mathrm{Cu}^{+}$ precipitates / is removed from the equilibrium / the concentration of copper(I) in solution is very low

- the equilibrium position moves to the right-hand

(1) side

Accept the actual electrode potential (for $\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}$ ) is higher/more positive than the standard electrode potential

Ignore reference to activation energy / rate of reaction/non-standard temperature/pressure Allow 1 mark for concentrations are non-standard If value in (i) for Ecell is positive then allow 1 mark in (ii) for reference to a spontaneous reaction for a positive Ecell

[Maximum number: 2]

This question is about manganese compounds. Some data are given below.

Table for Question 14 — Edexcel A-Level Chemistry A2

Calculate $E_{\text {cell }}^{\ominus}$ for the disproportionation of manganate(VI) ions in acidic conditions, stating whether or not the reaction is thermodynamically feasible.

selection of appropriate values and insertion in the correct equation (1)

- completion of calculation

and

evaluation of the feasibility of the reaction (1)

$E^{\text {o }}{ }_{\text {cell }}=2.26-0.56$ $=(+) 1.70(\mathrm{~V})$ and (positive so
reaction is) feasible
Accept (if second equation given)
$E^{\mathrm{o}}{ }_{\text {cell }}=1.51-0.56$ $=(+) 0.95(\mathrm{~V})$ and (positive so
TE on alkaline disproportionation
$E_{\text {cell }}^{\text {o }}=0.59-0.56=(+) 0.03(\mathrm{~V})$ and
(positive so reaction is) feasible scores (2)
TE on equations used the wrong way
round $-1.70 /-0.95(\mathrm{~V})$ and not feasible
scores (1)
Just (+)1.70 (V) / (+)0.95 (V) scores (1)
Do not award a non-disproportionation
reaction

[Maximum number: 5]

Brass is an alloy of copper. The percentage by mass of copper in a sample of brass can be determined by a three-stage process.

Stage 1 Oxidation of the copper to copper(II) ions with excess concentrated nitric acid.
Stage 2 Reduction of the copper(II) ions to copper(I) ions with excess iodide ions.
Stage 3 Titration of the iodine produced in Stage 2 against a standard solution of sodium thiosulfate.

(a)

The procedure in Stage 3 is shown:
- the iodine produced in each flask from Stage 2 is titrated with $0.095 \mathrm{~mol} \mathrm{dm}^{-3}$ sodium thiosulfate solution until the iodine colour is pale yellow
- a few drops of starch indicator are added
- sodium thiosulfate is added drop by drop with swirling until the end-point is reached
- the titration is repeated until two concordant titres are obtained.

[ 5 ]

(i)

The redox equation for the reaction between copper(II) ions and iodide ions is shown.

2 \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{aq})

The reaction of sodium thiosulfate solution with iodine is shown.

2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})

Calculate the percentage by mass of copper in the 2.53 g of brass.

\left[A_{r} \mathrm{Cu}=63.5\right]

[ 3 ]

calculation of moles thiosulfate in titre

- calculation of mass of copper in $25 \mathrm{~cm}^{3}$ as stoichiometry of Cu: thiosulfate is 1: 1

- calculation of % of copper by mass in sample

$\frac{26.55}{1000} \times 0.095=2.52225 \times 10^{-3} / 0.00252225(\mathrm{~mol})$ TE on (c)(iii)

$2.52(225) \times 10^{-3} \times 63.5=0.160163(\mathrm{~g})$ $\frac{0.160163}{2.53} \times 10 \times 100=63.305 / 63.31 / 63.3 / 63 \%$

TE on M2 unless greater than 100 %

Ignore SF except 1 SF

Correct answer with some working scores

(ii)

Nitric acid reacts with iodide ions to form iodine.

In another experiment the nitric acid was not neutralised before the start of Stage 2.

Explain the effect on the value obtained for the percentage of copper in the sample.

[ 2 ]

(iii)

Give a possible reason why the starch indicator is added when the iodine colour is pale yellow rather than at the start of the titration.

[ 1 ]

(iv)

Give the colour change at the end-point of the reaction.

[ 1 ]

[Maximum number: 4]

This question is about manganese compounds. Some data are given below.

Table for Question 16 — Edexcel A-Level Chemistry A2

(a)

The uncertainties of the apparatus used in the experiment in (b) are shown.

Table for Question (a) — Edexcel A-Level Chemistry A2

[ 4 ]

(i)

Complete the table.

[ 2 ]

$\text { (Balance: } \frac{100 \times 0.005 \times 2}{1.53}=0.65 \text { ) }$

- burette: $\frac{100 \times 0.05 \times 2}{27.35}=0.37(\%)$

- pipette: $\frac{100 \times 0.06}{25}=0.24(\%)$

- volumetric flask: $\frac{100 \times 0.3}{250}=0.12(\%)$

(Total =1.38 )

All three percentages correct scores (2)

Any two percentages correct scores (1)

IGNORE SF

(ii)

A third student obtained a value of 95.863% for the proportion of iron in the wire. State whether or not this student has given their answer to an appropriate number of significant figures. Justify your answer in terms of the total percentage uncertainty of the experiment.

[ 2 ]

[Maximum number: 1]

A hydrogen-oxygen fuel cell is used to provide electrical energy for an electric motor in a car.

Hydrogen-oxygen fuel cells have advantages over methanol-oxygen fuel cells in vehicles.

Which of these is an advantage of the hydrogen-oxygen fuel cell?

A

more energy is released per mole of fuel used

B

emissions do not contribute to climate change

C

hydrogen is easier to store than methanol

D

only hydrogen can be obtained from renewable resources

[Maximum number: 1]

A hydrogen-oxygen fuel cell is used to provide electrical energy for an electric motor in a car.

The electrolyte in the fuel cell is acidic. What is the half-equation at the anode?

A

$1 / 2 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

B

$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 1 / 2 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}$

C

$\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}$

D

$2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{~g})$

Edexcel A-Level Chemistry A2 16 Redox Equilibria Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question (v)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question 16

Question (a)

Question (i)

Question (ii)

Question 17

Question 18