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CHXII03:ELECTROCHEMISTRY

330086 A standard hydrogen electrode has a zero potential because

1 Hydrogen can be most easily oxidised

2 The electrode potential is assumed to be zero

3 Hydrogen has only one electron

4 Hydrogen is the lightest element

CHXII03:ELECTROCHEMISTRY

330087 Deduce from the following $E^{\circ}$ values of half cells, what combination of two half cells would result in a cell with the largest potential?
I. $A \to A^{+} + e; E^{\circ} = - 0.24 V$
II. $B^{+} \to B + e; E^{\circ} = - 2.1 V$
III. $C \to C^{2 +} + 2 e^{-}; E^{0} = - 0.38 V$
IV. $D^{2 -} \to D^{-} + e^{-}; E^{0} = - 0.59 V$

1 I and IV

2 II and III

3 III and IV

4 I and II

Explanation:

$E_{cell}^{o} = RHS - LHS = 0.24 - (- 2.1)$
$= 2.34 V$ for (i) & (ii) reactions.

CHXII03:ELECTROCHEMISTRY

330088 For the cell reaction, $2 C e^{4 +} + Co \to 2 C e^{3 +} C o^{2 +}; E_{cell}^{\circ} is 1.89 V .$ if $E_{C o^{2 +} / Co}$ is $- 0.28 V,$ What is the value of $E_{C e^{4 +} / C e^{3 +}}^{\circ}$ ?

1

0.28 V

2

1.61 V

3

2.17 V

4

5.29 V

Explanation:

$E {^{\circ}}_{C e l l} = E {^{\circ}}_{C e l l^{4 +} / C e^{3 +}} - E {^{\circ}}_{C o^{2 +} / C o}$
$\Rightarrow 1.89 = E_{C e^{4} / C e^{3 +}}^{\circ} - (- 0.28)$
$\Rightarrow E_{C e^{4 +} / C e^{3 +}}^{\circ} = 1.89 - 0.28 = 1.61 V$

CHXII03:ELECTROCHEMISTRY

330089 Which cell will measure standard electrode potential of copper electrode?

1

\begin{array}{l} P t (s) | H_{2} (g, 0.1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

2

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 2 M) | C u \end{array}

3

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

4

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 0 . 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

CHXII03:ELECTROCHEMISTRY

330086 A standard hydrogen electrode has a zero potential because

1 Hydrogen can be most easily oxidised

2 The electrode potential is assumed to be zero

3 Hydrogen has only one electron

4 Hydrogen is the lightest element

CHXII03:ELECTROCHEMISTRY

330087 Deduce from the following $E^{\circ}$ values of half cells, what combination of two half cells would result in a cell with the largest potential?
I. $A \to A^{+} + e; E^{\circ} = - 0.24 V$
II. $B^{+} \to B + e; E^{\circ} = - 2.1 V$
III. $C \to C^{2 +} + 2 e^{-}; E^{0} = - 0.38 V$
IV. $D^{2 -} \to D^{-} + e^{-}; E^{0} = - 0.59 V$

1 I and IV

2 II and III

3 III and IV

4 I and II

Explanation:

$E_{cell}^{o} = RHS - LHS = 0.24 - (- 2.1)$
$= 2.34 V$ for (i) & (ii) reactions.

CHXII03:ELECTROCHEMISTRY

330088 For the cell reaction, $2 C e^{4 +} + Co \to 2 C e^{3 +} C o^{2 +}; E_{cell}^{\circ} is 1.89 V .$ if $E_{C o^{2 +} / Co}$ is $- 0.28 V,$ What is the value of $E_{C e^{4 +} / C e^{3 +}}^{\circ}$ ?

1

0.28 V

2

1.61 V

3

2.17 V

4

5.29 V

Explanation:

$E {^{\circ}}_{C e l l} = E {^{\circ}}_{C e l l^{4 +} / C e^{3 +}} - E {^{\circ}}_{C o^{2 +} / C o}$
$\Rightarrow 1.89 = E_{C e^{4} / C e^{3 +}}^{\circ} - (- 0.28)$
$\Rightarrow E_{C e^{4 +} / C e^{3 +}}^{\circ} = 1.89 - 0.28 = 1.61 V$

CHXII03:ELECTROCHEMISTRY

330089 Which cell will measure standard electrode potential of copper electrode?

1

\begin{array}{l} P t (s) | H_{2} (g, 0.1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

2

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 2 M) | C u \end{array}

3

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

4

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 0 . 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

CHXII03:ELECTROCHEMISTRY

330086 A standard hydrogen electrode has a zero potential because

1 Hydrogen can be most easily oxidised

2 The electrode potential is assumed to be zero

3 Hydrogen has only one electron

4 Hydrogen is the lightest element

CHXII03:ELECTROCHEMISTRY

330087 Deduce from the following $E^{\circ}$ values of half cells, what combination of two half cells would result in a cell with the largest potential?
I. $A \to A^{+} + e; E^{\circ} = - 0.24 V$
II. $B^{+} \to B + e; E^{\circ} = - 2.1 V$
III. $C \to C^{2 +} + 2 e^{-}; E^{0} = - 0.38 V$
IV. $D^{2 -} \to D^{-} + e^{-}; E^{0} = - 0.59 V$

1 I and IV

2 II and III

3 III and IV

4 I and II

Explanation:

$E_{cell}^{o} = RHS - LHS = 0.24 - (- 2.1)$
$= 2.34 V$ for (i) & (ii) reactions.

CHXII03:ELECTROCHEMISTRY

330088 For the cell reaction, $2 C e^{4 +} + Co \to 2 C e^{3 +} C o^{2 +}; E_{cell}^{\circ} is 1.89 V .$ if $E_{C o^{2 +} / Co}$ is $- 0.28 V,$ What is the value of $E_{C e^{4 +} / C e^{3 +}}^{\circ}$ ?

1

0.28 V

2

1.61 V

3

2.17 V

4

5.29 V

Explanation:

$E {^{\circ}}_{C e l l} = E {^{\circ}}_{C e l l^{4 +} / C e^{3 +}} - E {^{\circ}}_{C o^{2 +} / C o}$
$\Rightarrow 1.89 = E_{C e^{4} / C e^{3 +}}^{\circ} - (- 0.28)$
$\Rightarrow E_{C e^{4 +} / C e^{3 +}}^{\circ} = 1.89 - 0.28 = 1.61 V$

CHXII03:ELECTROCHEMISTRY

330089 Which cell will measure standard electrode potential of copper electrode?

1

\begin{array}{l} P t (s) | H_{2} (g, 0.1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

2

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 2 M) | C u \end{array}

3

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

4

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 0 . 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

CHXII03:ELECTROCHEMISTRY

330086 A standard hydrogen electrode has a zero potential because

1 Hydrogen can be most easily oxidised

2 The electrode potential is assumed to be zero

3 Hydrogen has only one electron

4 Hydrogen is the lightest element

CHXII03:ELECTROCHEMISTRY

330087 Deduce from the following $E^{\circ}$ values of half cells, what combination of two half cells would result in a cell with the largest potential?
I. $A \to A^{+} + e; E^{\circ} = - 0.24 V$
II. $B^{+} \to B + e; E^{\circ} = - 2.1 V$
III. $C \to C^{2 +} + 2 e^{-}; E^{0} = - 0.38 V$
IV. $D^{2 -} \to D^{-} + e^{-}; E^{0} = - 0.59 V$

1 I and IV

2 II and III

3 III and IV

4 I and II

Explanation:

$E_{cell}^{o} = RHS - LHS = 0.24 - (- 2.1)$
$= 2.34 V$ for (i) & (ii) reactions.

CHXII03:ELECTROCHEMISTRY

330088 For the cell reaction, $2 C e^{4 +} + Co \to 2 C e^{3 +} C o^{2 +}; E_{cell}^{\circ} is 1.89 V .$ if $E_{C o^{2 +} / Co}$ is $- 0.28 V,$ What is the value of $E_{C e^{4 +} / C e^{3 +}}^{\circ}$ ?

1

0.28 V

2

1.61 V

3

2.17 V

4

5.29 V

Explanation:

$E {^{\circ}}_{C e l l} = E {^{\circ}}_{C e l l^{4 +} / C e^{3 +}} - E {^{\circ}}_{C o^{2 +} / C o}$
$\Rightarrow 1.89 = E_{C e^{4} / C e^{3 +}}^{\circ} - (- 0.28)$
$\Rightarrow E_{C e^{4 +} / C e^{3 +}}^{\circ} = 1.89 - 0.28 = 1.61 V$

CHXII03:ELECTROCHEMISTRY

330089 Which cell will measure standard electrode potential of copper electrode?

1

\begin{array}{l} P t (s) | H_{2} (g, 0.1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

2

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 2 M) | C u \end{array}

3

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

4

\begin{array}{l} P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q ., 0 . 1 M) | \\ C u^{2 +} (a q ., 1 M) | C u \end{array}

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