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Standard XII

Chemistry

EMF

The standard ...

Question

The standard emf for the given cell reaction, $Z n + {C u}^{2 +} ⟶ C u + {Z n}^{2 +}$ is $1.10 V$ at $25^{o} C$ . The emf for the cell reaction, when $0.1 M$ ${C u}^{2 +}$ and $0.1 M$ ${Z n}^{2 +}$ solutions are used at $25^{o} C$ is:

1.10 V

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- 1.10 V

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2.20 V

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- 2.20 V

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Solution

The correct option is B $1.10 V$
$Z n + C u^{2 +} ⟶ Z n^{2 +} + C u$

From Nernst equation:

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} log \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]}$

$E_{c e l l} = 1.10 - \frac{0.059}{2} log \frac{0.1}{0.1} = 1.10 V$

Hence, option A is correct.

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Similar questions

Q. The standard cell potential for

Z n | Z n^{2 +} | | C u^{2 +} | C u

is 1.10V. When the cell is completely discharged,

l o g [Z n^{2 +}] / [C u^{2 +}]

is closest to:

Q. For the reaction,

Z n (s) + {C u}^{2 +} (0.1 M) ⟶ {Z n}^{2 +} (1 M) + C u (s)

taking place in a cell;

E_{c e l l}^{o}

1.10 V

E_{c e l l}

for the cell will be

(2.303 \frac{R T}{F} = 0.0591)

Q. Given that (at

T = 298 K)

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s) E_{c e l l}^{\circ} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s) E_{c e l l}^{\circ} = 1.10 V

The

E_{c e l l}

for

Z n | Z n^{2 +} (0.1 M) | | A g^{+} (1.0 M | A g

298 K

will be:

E^{\circ}

for the electrochemical cell,

Z n (s) | Z n^{2 +} 1 M (a q .) ∥ C u^{2 +} 1 M (a q .) | C u (s)

1.10 V

25^{\circ} C

. The equilibrium constant for the cell reaction,

Z n (s) + C u^{2 +} (a q .) ⇌ Z n^{2 +} (a q .) + C u (s)

will be:

$E^{0}$ for the cell, $Z n | Z n {^{2 +}}_{(a q)} | | C u {^{2 +}}_{(a q)} | C u$ is $1.10 V$ at $25^{0} C$ . The equilibrium constant for $Z n + C u {^{2 +}}_{(a q)} \to Z n {^{2 +}}_{(a q)} + C u$ the cell reaction is of the order of :

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The standard emf for the given cell reaction, Zn+Cu2+⟶Cu+Zn2+ is 1.10 V at 25oC. The emf for the cell reaction, when 0.1M Cu2+ and 0.1M Zn2+ solutions are used at 25oC is:

The correct option is B 1.10 VZn+Cu2+⟶Zn2++CuFrom Nernst equation:Ecell=E0cell−0.059nlog[Zn2+][Cu2+]Ecell=1.10−0.0592log0.10.1 =1.10 VHence, option A is correct.

The standard emf for the given cell reaction, $Z n + {C u}^{2 +} ⟶ C u + {Z n}^{2 +}$ is $1.10 V$ at $25^{o} C$ . The emf for the cell reaction, when $0.1 M$ ${C u}^{2 +}$ and $0.1 M$ ${Z n}^{2 +}$ solutions are used at $25^{o} C$ is:

The correct option is B $1.10 V$
$Z n + C u^{2 +} ⟶ Z n^{2 +} + C u$

From Nernst equation:

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} log \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]}$

$E_{c e l l} = 1.10 - \frac{0.059}{2} log \frac{0.1}{0.1} = 1.10 V$

Hence, option A is correct.