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Electrolysis and Electrolytes

Zn s + Cu 2+0...

Question

$Z n (s) + C u^{2 +} (0.1 M) \to Z n^{2 +} (1 M) + C u (s)$

The reaction is taking place in cell, where $E_{c e l l}^{0}$ is 1.10 volt; $E_{c e l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

2.14 volt

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1.80 volt

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1.07 volt

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0.82 volt

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Solution

The correct option is C 1.07 volt
Explanation :

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$= 1.10 - \frac{0.0591}{2} l o g [10]$

$= 1.10 - 0.0295 = 1.07 V$

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Similar questions

For the redox reaction $Z n (S) + C u^{2 +} (0.1 M) \to Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{c e l l}^{0}$ is 1.10 volt. $E_{c e l l}$ for the cell will be : $(2.303 \frac{R T}{F} = 0.0591)$

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. For the cell,

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

The cell potential

E_{c e l l}

can be increased:

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

Reaction quotient

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

, variation

E_{c e l l}

with

Q

is of the type with

O A = 1.10 v o l t . E_{c e l l} = 1.1591 v o l t

when:

Q. For the redox reaction,

Z n + C u_{(0.1 M)}^{2 +} \to Z n_{(1 M)}^{2 +} + C u

taking place in a cell

E_{c e l l}^{⊖} = 1.10

E_{C e l l}^{⊖}

for the galvanic cell will be?

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Zn(s)+Cu2+(0.1M)→Zn2+(1M)+Cu(s) The reaction is taking place in cell, where E0cell is 1.10 volt; Ecell for the cell will be (2.303RTF=0.0591)

The correct option is C 1.07 voltExplanation : Ecell=E0cell−0.0591nlog[Zn2+][Cu2+] =1.10−0.05912log[10] =1.10−0.0295=1.07 V

$Z n (s) + C u^{2 +} (0.1 M) \to Z n^{2 +} (1 M) + C u (s)$

The reaction is taking place in cell, where $E_{c e l l}^{0}$ is 1.10 volt; $E_{c e l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

The correct option is C 1.07 volt
Explanation :

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$= 1.10 - \frac{0.0591}{2} l o g [10]$

$= 1.10 - 0.0295 = 1.07 V$