For the cell- Zn-Zn2- 1 M-H-1 M- PtH2- 1 bar- E-cell - 0-74 V and for the cell PtH2-H-1 M-Ag-1 M-Ag- E-cell - 0-80 VThus- the value of Ecell and spontaneity for the following reaction respectively areAg-Ag- 0-1 M-Zn2-0-1 M- Zn

For the cell, Zn|Zn^2+ (1 M)||H^+(1 M)|Pt(H_2, 1 bar) E^∘_cell = nbsp; E^∘_SHE E^∘_Zn^2+/Zn 0.74 = 0.00 E^∘_Zn^2+/Zn ∴ E^∘_Zn^2+/Zn = 0.74 V Similarly, w ...

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Byju's Answer

Standard VII

Chemistry

Equilibrium Constant from Nernst Equation

For the cell,...

Question

For the cell, $Z n | Z n^{2 +} (1 M) | | H^{+} (1 M) | P t (H_{2}, 1 b a r), E_{c e l l}^{\circ} = 0.74 V$ and for the cell $P t (H_{2}) | H^{+} (1 M) | | A g^{+} (1 M) | A g, E_{c e l l}^{\circ} = 0.80 V$
Thus, the value of $E_{c e l l}$ and spontaneity for the following reaction respectively are
$A g | A g^{+} (0.1 M) | | Z n^{2 +} (0.1 M) | Z n$

1.44 V

and spontaneous

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0.4 V

and spontaneous

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- 1.44 V

and non-spontaneous

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- 1.53 V

and non-spontaneous

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Solution

The correct option is D $- 1.53 V$ and non-spontaneous
For the cell,

$Z n | Z n^{2 +} (1 M) | | H^{+} (1 M) | P t (H_{2}, 1 b a r)$

$E_{c e l l}^{\circ} = E_{S H E}^{\circ} - E_{Z n^{2 +} / Z n}^{\circ}$
$0.74 = 0.00 - E_{Z n^{2 +}}^{\circ} / Z n$
$∴ E_{Z n^{2 +} / Z n}^{\circ} = - 0.74 V$
Similarly, we can calculate the value of $E_{A g^{+} / A g}^{\circ}$ :
$P t (H_{2}, 1 b a r) | H^{+} (1 M) | | A g^{+} (1 M) | A g$
$E_{c e l l}^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{S H E}^{\circ}$
$\Rightarrow E_{A g^{+} / A g}^{\circ} = 0.80 V$

Now, for the cell
$A g | A g^{+} (0.1 M) | | Z n^{2 +} (0.1 M) | Z n$
$E_{c e l l}^{\circ} = E_{Z n^{2 +} / Z n}^{\circ} - E_{A g^{+} / A g}^{\circ} = - 0.74 V - (0.80 V) = - 1.54 V$
Reaction quotient $(Q) = \frac{[A g^{+}]^{2}}{[Z n^{2 +}]} = \frac{(0.1)^{2}}{(0.1)} = \frac{0.01}{0.1} = \frac{1}{10}$
$∴ E_{c e l l} = E_{c e l l}^{\circ} - \frac{2.303 R T}{n F} l o g Q$
At, $298 K, \frac{2.303 R T}{F} = 0.0591 V$
$∴ E_{c e l l} = - 1.54 - \frac{0.0591}{2} l o g (\frac{1}{10})$
$= - 1.54 + \frac{0.0591}{2} l o g 10$
$= - 1.54 + 0.0295$
$= - 1.51 V$
$E_{c e l l} < 0$ , hence reaction is non-spontaneous.
So, (d) is correct.

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Similar questions

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)

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:

Q. Calculate the emf of the following cell at

298 K

Z n (s) ∣ Z n^{2 +} (a q, 0.1 M) ∥ A g^{+} (a q, 0.01 M) ∣ A g (s)

Given:

E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

E_{A g^{+} / A g}^{0} = + 0.80 V

Q. The emf of the following cells are:

A g | A g^{+} (1 M) | | C u^{2 +} (1 M) | C u E_{c e l l} = - 0.46 V a n d

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u E_{c e l l} = + 1.10 V .

Then the emf of the cell

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

is:

Q. The emf

(E^{\circ})

of the following cells are:

A g / A g^{+} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = - 0.46 v o l t

Z n / Z n^{2 +} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = + 1.10 v o l t

Calculate the emf of the cell:

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

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For the cell, Zn|Zn2+(1 M)||H+(1 M)|Pt(H2, 1 bar),E∘cell=0.74 V and for the cell Pt(H2)|H+(1 M)||Ag+(1 M)|Ag,E∘cell=0.80 V Thus, the value of Ecell and spontaneity for the following reaction respectively are Ag|Ag+(0.1 M)||Zn2+(0.1 M)|Zn