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Byju's Answer

Standard XII

Chemistry

Nernst Equation

For the react...

Question

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; $E_{c e l l}^{o}$ is $1.10 V$ . $E_{c e l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

1.80 V

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1.07 V

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0.82 V

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2.14 V

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Solution

The correct option is D $1.07 V$
From the reaction, we get

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

$= 1.10 - \frac{0.0591}{2} {log}_{10} \frac{1}{0.1} = 1.07 V$

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

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)

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:

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?

Q. At

298 K

, given that:

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s)

E_{c e l l}^{o} = 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}^{o} = 1.10 V

Then, the

E_{c e l l}

for the following reaction at

298 K

will be:

Z n (s) | Z n^{2 +} (0.1 M) | | A g^{+} (1. 0 M) | A g (s)

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For the reaction, Zn(s)+Cu2+(0.1M)⟶Zn2+(1M)+Cu(s) taking place in a cell; Eocell is 1.10V. Ecell for the cell will be (2.303RTF=0.0591)

The correct option is D 1.07VFrom the reaction, we getEcell=Eocell−0.0591nlog10[Zn2+][Cu2+] =1.10−0.05912log1010.1=1.07V

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; $E_{c e l l}^{o}$ is $1.10 V$ . $E_{c e l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

The correct option is D $1.07 V$
From the reaction, we get

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

$= 1.10 - \frac{0.0591}{2} {log}_{10} \frac{1}{0.1} = 1.07 V$