For the galvanic cell- Zn s -Cu2- 0-02 M - Zn2- 0-04 M -Cu s Ecell- -10-2 V Nearest integer- Use - E0Cu-Cu2- -0-34 V - E0Zn-Zn2-0-76 V-2-303 RTF-0-059 V -

The standard reduction potential of the half cells are E^0_Cu^2+/Cu= +0.34 V E^0_Zn^2+/Zn= 0.76 V Zn ( s )+Cu^2+ ( 0.02 M )→ Zn^2+ ( 0.04 M )+Cu ( s ) In ...

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Byju's Answer

Standard XII

Chemistry

Electrolytic Cell

For the galva...

Question

For the galvanic cell,

$Z n (s) + C u^{2 +} (0.02 M) \to Z n^{2 +} (0.04 M) + C u (s)$

$E_{c e l l} =$ _______ $\times 10^{- 2} V$ (Nearest integer)

$[U s e : E_{C u / C u^{2 +}}^{0} = - 0.34 V, E_{Z n / Z n^{2 +}}^{0} = + 0.76 V, \frac{2.303 R T}{F} = 0.059 V]$

109.00

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109.0

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109

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Solution

The standard reduction potential of the half cells are
$E_{C u^{2 +} / C u}^{0} = + 0.34 V$
$E_{Z n^{2 +} / Z n}^{0} = - 0.76 V$

$Z n (s) + C u^{2 +} (0.02 M) \to Z n^{2 +} (0.04 M) + C u (s)$

In the given reaction, $C u^{2 +}$ get reduced and $Z n$ get oxidised.
Hence,
$E_{c e l l}^{0} = E_{c a t h o d e (r e d)}^{0} - E_{a n o d e (r e d)}^{0}$
$E_{c e l l}^{0} = 0.34 - (- 0.76)$
$E_{c e l l}^{0} = + 1.1 V$

$R e a c t i o n q u o t i e n t, Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]} = \frac{0.04}{0.02} = 2$

$E_{c e l l} = E_{c e l l}^{0} + \frac{0.059}{n} l o g Q$
n is number of electron involved in the reaction.

$E_{c e l l} = 1.1 - \frac{0.059}{2} l o g 2$

$= 109 \times 10^{- 2} V$

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Similar questions

Q. The standard reduction potential

E^{\circ}

, for the half reaction are:-

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

The emf for the cell reaction,

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

is_____________.

Q. Calculate the emf of the cell.

Z n | Z n^{2 +} (0.001 M) | | C u^{2 +} (0.1 M) | C u

The standard potential of

C u / C u^{2 +}

half-cell is +0.34 and

Z n / Z n^{2 +}

is -0.76 V.

Consider the cell:

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

The standard reduction potential are 0.350 V for

C u^{2 +} (a q) + 2 e^{-} \to C u

and -0.763 V for

Z n^{2 +} (a q) + 2 e^{-} \to Z n

The EMF (in V) of the cell is ( write your answer as nearest integer) :

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

E^{o}

for

C u

and

Z n

are

0.35

and

- 0.76

respectively:

Find EMF of cell (nearest integer) if

Z n^{2 +}

and

C u^{2 +}

are 1 M each.

The standard reducation potentials of

$Z n^{2 +} | Z n, C u^{2 +} | C u$ and $A g^{+} | A g$

are respectively -0.76, 0.34 and 0.8 V. The following cells were constructed.

a) $Z n | Z n^{2 +} | | C u^{2 +} | C u$

b) $Z n | Z n^{2 +} | | A g^{+} | A g$

c) $C u | C u^{2 +} | | A g^{+} | A g$

What is the correct order $E {^{0}}_{c e l l}$ of these cells?

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For the galvanic cell, Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s) Ecell= _______ ×10−2 V (Nearest integer) [Use : E0Cu/Cu2+=−0.34 V, E0Zn/Zn2+=+0.76 V,2.303 RTF=0.059 V]

For the galvanic cell,

$Z n (s) + C u^{2 +} (0.02 M) \to Z n^{2 +} (0.04 M) + C u (s)$

$E_{c e l l} =$ _______ $\times 10^{- 2} V$ (Nearest integer)

$[U s e : E_{C u / C u^{2 +}}^{0} = - 0.34 V, E_{Z n / Z n^{2 +}}^{0} = + 0.76 V, \frac{2.303 R T}{F} = 0.059 V]$