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CHXII03:ELECTROCHEMISTRY

329986 The potential of hydrogen electrode is $- 118 m V$ . The $H^{+}$ concentration of the solution is

1

1 0^{- 4} M

2

2 M

3

0 .01 M

4

1 M

Explanation:

$- 118 m V = - 0 .118 V$
$E = \frac{0 .0591}{n} l o g [H^{+}]$
$- 0 .118 = \frac{0 .059}{1} l o g [H^{+}]$
$l o g [H^{+}] = - 2$
$[H^{+}] = 1 0^{- 2}$

CHXII03:ELECTROCHEMISTRY

329987 Emf of the cell
$N i | N i^{2 +} (0 .1 M) | A u^{3 +} (1 .0 M) | A u$ will be
$E_{N i | N i^{2 +}}^{^{\circ}} = 0 .25, E_{A u | A u^{3 +}}^{^{\circ}} = - 0 .15 V$

1

- 1 .7795 V

2

+ 1 .7795 V

3

1 .75 V

4

+ 0 .7795 V

Explanation:

Cell reaction :
$3 N i + 2 A u^{+ 3} \to 3 N i^{+ 2} + 2 A u$
$E_{c e l l} {= E}_{c e l l}^{^{\circ}} - \frac{0 .0591}{6} l o g \frac{{[N i^{+ 2}]}^{3}}{{[A u^{+ 3}]}^{2}}$
$= (0 .25 + 1 .5) - \frac{0 .0591}{6} l o g \frac{{(0 .1)}^{3}}{{(1)}^{2}}$
$= 1 .75 + \frac{0 .0591}{2}$
$= 1 .75 + 0 .295 = + 1 .7795 V$

CHXII03:ELECTROCHEMISTRY

329988 The cell,
$Z n | Z n^{2 +} (1 M) | C u^{2 +} (1 M) | {C u (E}_{c e l l}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +} t o C u^{2 +}, [\frac{Z n^{2 +}}{C u^{2 +}}]$ is

1

9 .65 \times 1 0^{4}

2

1 0^{37 .3}

3 antilog (24.08)

4 37.3

Explanation:

$E_{c e l l} {= E}_{c e l l}^{o} - \frac{0 .0591}{n} l o g Q$
$W h e r e, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
For complete discharge, $E_{c e l l} = 0$
$S o, E_{c e l l}^{o} = \frac{0 .0591}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$∴ [\frac{[Z n^{2 +}]}{[C u^{2 +}]}] = 1 0^{37 .3}$

CHXII03:ELECTROCHEMISTRY

329989 The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is

1 Greater than 1

2 Less than 1

3 Equal to 1

4 None of the above

Explanation:

$E = E^{o} - \frac{0 .0591}{n} l o g \frac{[P r o d u c t]}{[R e a c t a n t]}$
$i f \frac{[P r o d u c t]}{[R e a c t a n t]} = 1, t h e n E = E^{o}$

CHXII03:ELECTROCHEMISTRY

329990 The oxidation potential of $0 .05 M H_{2} S O_{4}$ is

1

- 2 \times 0 .0591

2

- 0 .01 \times 0 .0591

3

- 2 .321 \times 0 .0591

4

0 .0591

Explanation:

$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{R e d} {= E}_{R e d}^{o} - \frac{0 .059}{n} l o g \frac{1}{{[H^{+}]}^{- 2}};$
$E_{R e d} = 0 . - \frac{0 .059}{2} l o g \frac{1}{{(0 .1)}^{2}}; E_{R e d} = - 0 .059 V,$
$E_{a x i} = 0 .059 V .$

CHXII03:ELECTROCHEMISTRY

329986 The potential of hydrogen electrode is $- 118 m V$ . The $H^{+}$ concentration of the solution is

1

1 0^{- 4} M

2

2 M

3

0 .01 M

4

1 M

Explanation:

$- 118 m V = - 0 .118 V$
$E = \frac{0 .0591}{n} l o g [H^{+}]$
$- 0 .118 = \frac{0 .059}{1} l o g [H^{+}]$
$l o g [H^{+}] = - 2$
$[H^{+}] = 1 0^{- 2}$

CHXII03:ELECTROCHEMISTRY

329987 Emf of the cell
$N i | N i^{2 +} (0 .1 M) | A u^{3 +} (1 .0 M) | A u$ will be
$E_{N i | N i^{2 +}}^{^{\circ}} = 0 .25, E_{A u | A u^{3 +}}^{^{\circ}} = - 0 .15 V$

1

- 1 .7795 V

2

+ 1 .7795 V

3

1 .75 V

4

+ 0 .7795 V

Explanation:

Cell reaction :
$3 N i + 2 A u^{+ 3} \to 3 N i^{+ 2} + 2 A u$
$E_{c e l l} {= E}_{c e l l}^{^{\circ}} - \frac{0 .0591}{6} l o g \frac{{[N i^{+ 2}]}^{3}}{{[A u^{+ 3}]}^{2}}$
$= (0 .25 + 1 .5) - \frac{0 .0591}{6} l o g \frac{{(0 .1)}^{3}}{{(1)}^{2}}$
$= 1 .75 + \frac{0 .0591}{2}$
$= 1 .75 + 0 .295 = + 1 .7795 V$

CHXII03:ELECTROCHEMISTRY

329988 The cell,
$Z n | Z n^{2 +} (1 M) | C u^{2 +} (1 M) | {C u (E}_{c e l l}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +} t o C u^{2 +}, [\frac{Z n^{2 +}}{C u^{2 +}}]$ is

1

9 .65 \times 1 0^{4}

2

1 0^{37 .3}

3 antilog (24.08)

4 37.3

Explanation:

$E_{c e l l} {= E}_{c e l l}^{o} - \frac{0 .0591}{n} l o g Q$
$W h e r e, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
For complete discharge, $E_{c e l l} = 0$
$S o, E_{c e l l}^{o} = \frac{0 .0591}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$∴ [\frac{[Z n^{2 +}]}{[C u^{2 +}]}] = 1 0^{37 .3}$

CHXII03:ELECTROCHEMISTRY

329989 The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is

1 Greater than 1

2 Less than 1

3 Equal to 1

4 None of the above

Explanation:

$E = E^{o} - \frac{0 .0591}{n} l o g \frac{[P r o d u c t]}{[R e a c t a n t]}$
$i f \frac{[P r o d u c t]}{[R e a c t a n t]} = 1, t h e n E = E^{o}$

CHXII03:ELECTROCHEMISTRY

329990 The oxidation potential of $0 .05 M H_{2} S O_{4}$ is

1

- 2 \times 0 .0591

2

- 0 .01 \times 0 .0591

3

- 2 .321 \times 0 .0591

4

0 .0591

Explanation:

$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{R e d} {= E}_{R e d}^{o} - \frac{0 .059}{n} l o g \frac{1}{{[H^{+}]}^{- 2}};$
$E_{R e d} = 0 . - \frac{0 .059}{2} l o g \frac{1}{{(0 .1)}^{2}}; E_{R e d} = - 0 .059 V,$
$E_{a x i} = 0 .059 V .$

CHXII03:ELECTROCHEMISTRY

329986 The potential of hydrogen electrode is $- 118 m V$ . The $H^{+}$ concentration of the solution is

1

1 0^{- 4} M

2

2 M

3

0 .01 M

4

1 M

Explanation:

$- 118 m V = - 0 .118 V$
$E = \frac{0 .0591}{n} l o g [H^{+}]$
$- 0 .118 = \frac{0 .059}{1} l o g [H^{+}]$
$l o g [H^{+}] = - 2$
$[H^{+}] = 1 0^{- 2}$

CHXII03:ELECTROCHEMISTRY

329987 Emf of the cell
$N i | N i^{2 +} (0 .1 M) | A u^{3 +} (1 .0 M) | A u$ will be
$E_{N i | N i^{2 +}}^{^{\circ}} = 0 .25, E_{A u | A u^{3 +}}^{^{\circ}} = - 0 .15 V$

1

- 1 .7795 V

2

+ 1 .7795 V

3

1 .75 V

4

+ 0 .7795 V

Explanation:

Cell reaction :
$3 N i + 2 A u^{+ 3} \to 3 N i^{+ 2} + 2 A u$
$E_{c e l l} {= E}_{c e l l}^{^{\circ}} - \frac{0 .0591}{6} l o g \frac{{[N i^{+ 2}]}^{3}}{{[A u^{+ 3}]}^{2}}$
$= (0 .25 + 1 .5) - \frac{0 .0591}{6} l o g \frac{{(0 .1)}^{3}}{{(1)}^{2}}$
$= 1 .75 + \frac{0 .0591}{2}$
$= 1 .75 + 0 .295 = + 1 .7795 V$

CHXII03:ELECTROCHEMISTRY

329988 The cell,
$Z n | Z n^{2 +} (1 M) | C u^{2 +} (1 M) | {C u (E}_{c e l l}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +} t o C u^{2 +}, [\frac{Z n^{2 +}}{C u^{2 +}}]$ is

1

9 .65 \times 1 0^{4}

2

1 0^{37 .3}

3 antilog (24.08)

4 37.3

Explanation:

$E_{c e l l} {= E}_{c e l l}^{o} - \frac{0 .0591}{n} l o g Q$
$W h e r e, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
For complete discharge, $E_{c e l l} = 0$
$S o, E_{c e l l}^{o} = \frac{0 .0591}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$∴ [\frac{[Z n^{2 +}]}{[C u^{2 +}]}] = 1 0^{37 .3}$

CHXII03:ELECTROCHEMISTRY

329989 The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is

1 Greater than 1

2 Less than 1

3 Equal to 1

4 None of the above

Explanation:

$E = E^{o} - \frac{0 .0591}{n} l o g \frac{[P r o d u c t]}{[R e a c t a n t]}$
$i f \frac{[P r o d u c t]}{[R e a c t a n t]} = 1, t h e n E = E^{o}$

CHXII03:ELECTROCHEMISTRY

329990 The oxidation potential of $0 .05 M H_{2} S O_{4}$ is

1

- 2 \times 0 .0591

2

- 0 .01 \times 0 .0591

3

- 2 .321 \times 0 .0591

4

0 .0591

Explanation:

$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{R e d} {= E}_{R e d}^{o} - \frac{0 .059}{n} l o g \frac{1}{{[H^{+}]}^{- 2}};$
$E_{R e d} = 0 . - \frac{0 .059}{2} l o g \frac{1}{{(0 .1)}^{2}}; E_{R e d} = - 0 .059 V,$
$E_{a x i} = 0 .059 V .$

CHXII03:ELECTROCHEMISTRY

329986 The potential of hydrogen electrode is $- 118 m V$ . The $H^{+}$ concentration of the solution is

1

1 0^{- 4} M

2

2 M

3

0 .01 M

4

1 M

Explanation:

$- 118 m V = - 0 .118 V$
$E = \frac{0 .0591}{n} l o g [H^{+}]$
$- 0 .118 = \frac{0 .059}{1} l o g [H^{+}]$
$l o g [H^{+}] = - 2$
$[H^{+}] = 1 0^{- 2}$

CHXII03:ELECTROCHEMISTRY

329987 Emf of the cell
$N i | N i^{2 +} (0 .1 M) | A u^{3 +} (1 .0 M) | A u$ will be
$E_{N i | N i^{2 +}}^{^{\circ}} = 0 .25, E_{A u | A u^{3 +}}^{^{\circ}} = - 0 .15 V$

1

- 1 .7795 V

2

+ 1 .7795 V

3

1 .75 V

4

+ 0 .7795 V

Explanation:

Cell reaction :
$3 N i + 2 A u^{+ 3} \to 3 N i^{+ 2} + 2 A u$
$E_{c e l l} {= E}_{c e l l}^{^{\circ}} - \frac{0 .0591}{6} l o g \frac{{[N i^{+ 2}]}^{3}}{{[A u^{+ 3}]}^{2}}$
$= (0 .25 + 1 .5) - \frac{0 .0591}{6} l o g \frac{{(0 .1)}^{3}}{{(1)}^{2}}$
$= 1 .75 + \frac{0 .0591}{2}$
$= 1 .75 + 0 .295 = + 1 .7795 V$

CHXII03:ELECTROCHEMISTRY

329988 The cell,
$Z n | Z n^{2 +} (1 M) | C u^{2 +} (1 M) | {C u (E}_{c e l l}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +} t o C u^{2 +}, [\frac{Z n^{2 +}}{C u^{2 +}}]$ is

1

9 .65 \times 1 0^{4}

2

1 0^{37 .3}

3 antilog (24.08)

4 37.3

Explanation:

$E_{c e l l} {= E}_{c e l l}^{o} - \frac{0 .0591}{n} l o g Q$
$W h e r e, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
For complete discharge, $E_{c e l l} = 0$
$S o, E_{c e l l}^{o} = \frac{0 .0591}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$∴ [\frac{[Z n^{2 +}]}{[C u^{2 +}]}] = 1 0^{37 .3}$

CHXII03:ELECTROCHEMISTRY

329989 The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is

1 Greater than 1

2 Less than 1

3 Equal to 1

4 None of the above

Explanation:

$E = E^{o} - \frac{0 .0591}{n} l o g \frac{[P r o d u c t]}{[R e a c t a n t]}$
$i f \frac{[P r o d u c t]}{[R e a c t a n t]} = 1, t h e n E = E^{o}$

CHXII03:ELECTROCHEMISTRY

329990 The oxidation potential of $0 .05 M H_{2} S O_{4}$ is

1

- 2 \times 0 .0591

2

- 0 .01 \times 0 .0591

3

- 2 .321 \times 0 .0591

4

0 .0591

Explanation:

$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{R e d} {= E}_{R e d}^{o} - \frac{0 .059}{n} l o g \frac{1}{{[H^{+}]}^{- 2}};$
$E_{R e d} = 0 . - \frac{0 .059}{2} l o g \frac{1}{{(0 .1)}^{2}}; E_{R e d} = - 0 .059 V,$
$E_{a x i} = 0 .059 V .$

CHXII03:ELECTROCHEMISTRY

329986 The potential of hydrogen electrode is $- 118 m V$ . The $H^{+}$ concentration of the solution is

1

1 0^{- 4} M

2

2 M

3

0 .01 M

4

1 M

Explanation:

$- 118 m V = - 0 .118 V$
$E = \frac{0 .0591}{n} l o g [H^{+}]$
$- 0 .118 = \frac{0 .059}{1} l o g [H^{+}]$
$l o g [H^{+}] = - 2$
$[H^{+}] = 1 0^{- 2}$

CHXII03:ELECTROCHEMISTRY

329987 Emf of the cell
$N i | N i^{2 +} (0 .1 M) | A u^{3 +} (1 .0 M) | A u$ will be
$E_{N i | N i^{2 +}}^{^{\circ}} = 0 .25, E_{A u | A u^{3 +}}^{^{\circ}} = - 0 .15 V$

1

- 1 .7795 V

2

+ 1 .7795 V

3

1 .75 V

4

+ 0 .7795 V

Explanation:

Cell reaction :
$3 N i + 2 A u^{+ 3} \to 3 N i^{+ 2} + 2 A u$
$E_{c e l l} {= E}_{c e l l}^{^{\circ}} - \frac{0 .0591}{6} l o g \frac{{[N i^{+ 2}]}^{3}}{{[A u^{+ 3}]}^{2}}$
$= (0 .25 + 1 .5) - \frac{0 .0591}{6} l o g \frac{{(0 .1)}^{3}}{{(1)}^{2}}$
$= 1 .75 + \frac{0 .0591}{2}$
$= 1 .75 + 0 .295 = + 1 .7795 V$

CHXII03:ELECTROCHEMISTRY

329988 The cell,
$Z n | Z n^{2 +} (1 M) | C u^{2 +} (1 M) | {C u (E}_{c e l l}^{o} = 1 .10 V)$ , was allowed to be completely discharged at 298 K. The relative concentration of $Z n^{2 +} t o C u^{2 +}, [\frac{Z n^{2 +}}{C u^{2 +}}]$ is

1

9 .65 \times 1 0^{4}

2

1 0^{37 .3}

3 antilog (24.08)

4 37.3

Explanation:

$E_{c e l l} {= E}_{c e l l}^{o} - \frac{0 .0591}{n} l o g Q$
$W h e r e, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
For complete discharge, $E_{c e l l} = 0$
$S o, E_{c e l l}^{o} = \frac{0 .0591}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$∴ [\frac{[Z n^{2 +}]}{[C u^{2 +}]}] = 1 0^{37 .3}$

CHXII03:ELECTROCHEMISTRY

329989 The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is

1 Greater than 1

2 Less than 1

3 Equal to 1

4 None of the above

Explanation:

$E = E^{o} - \frac{0 .0591}{n} l o g \frac{[P r o d u c t]}{[R e a c t a n t]}$
$i f \frac{[P r o d u c t]}{[R e a c t a n t]} = 1, t h e n E = E^{o}$

CHXII03:ELECTROCHEMISTRY

329990 The oxidation potential of $0 .05 M H_{2} S O_{4}$ is

1

- 2 \times 0 .0591

2

- 0 .01 \times 0 .0591

3

- 2 .321 \times 0 .0591

4

0 .0591

Explanation:

$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{R e d} {= E}_{R e d}^{o} - \frac{0 .059}{n} l o g \frac{1}{{[H^{+}]}^{- 2}};$
$E_{R e d} = 0 . - \frac{0 .059}{2} l o g \frac{1}{{(0 .1)}^{2}}; E_{R e d} = - 0 .059 V,$
$E_{a x i} = 0 .059 V .$

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