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Byju's Answer

Standard XII

Chemistry

Redox Reactions and Couple

The standard ...

Question

The standard cell potential for the cell $Z n ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ ∣ ∣ {C u}^{2 +} (1 M) ∣ ∣ C u$
gien $E_{{C u}^{2 +} / C u}^{o} = 0.34 V$
and $E_{{z n}^{2 +} / Z n}^{o} = - 0.76 V$ is

1.10 V

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- 0.42 V

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- 1.10 V

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None of these

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Solution

The correct option is A $1.10 V$
$E_{c e l l}^{o} = E_{c a t h o d e}^{o} + E_{a n o d e}^{o}$
$= E_{{C u}^{2 +} / C u}^{o} - E_{{z n}^{2 +} / Z n}^{o}$
$= (0.34 - (- 0.76)$
$= 0.34 + 0.76$
$E_{c e l l}^{o} = + 1.10 V$

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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. The EMFs of the following cells are

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{o} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{o} = 1.10 V

The EMF of the cell

Z n | Z n^{2 +} | | A g^{+} (1 M) | A g

will be:

Q. The EMF of the following cells are:

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{0} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{0} = 1.10 V

The EMF of the cell

Z n | Z n^{2 +} (1 M) | | A g^{+} (1 M) | A g

will be:

The standard cell potential for $Z n | Z n^{2 +} | | C u^{2 +} | C u i s 1.10 V .$ When the cell is completely discharged, $l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$ is closest to

Q. The cell,

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u

, was allowed to be completely discharged at 298 K. The relative concentration of

Z n^{2 +}

C u^{2 +} (\frac{[Z n^{2 +}]}{[C u^{2 +}]})

is :

[E_{c e l l}^{0} = 1.10 V]

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The standard cell potential for the cell Zn∣∣Zn2+(1M)∣∣∣∣Cu2+(1M)∣∣Cugien EoCu2+/Cu=0.34Vand Eozn2+/Zn=−0.76V is

The correct option is A 1.10VEocell=Eocathode+Eoanode=EoCu2+/Cu−Eozn2+/Zn=(0.34−(−0.76)=0.34+0.76Eocell=+1.10V

The standard cell potential for the cell $Z n ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ ∣ ∣ {C u}^{2 +} (1 M) ∣ ∣ C u$
gien $E_{{C u}^{2 +} / C u}^{o} = 0.34 V$
and $E_{{z n}^{2 +} / Z n}^{o} = - 0.76 V$ is

The correct option is A $1.10 V$
$E_{c e l l}^{o} = E_{c a t h o d e}^{o} + E_{a n o d e}^{o}$
$= E_{{C u}^{2 +} / C u}^{o} - E_{{z n}^{2 +} / Z n}^{o}$
$= (0.34 - (- 0.76)$
$= 0.34 + 0.76$
$E_{c e l l}^{o} = + 1.10 V$