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Gibb's Free Energy

A galvanic ce...

Question

A galvanic cell is formed by immersing $A g (s)$ in $A g N O_{3} (a q, 0.1 M)$ and $C u (s)$ in $C u S O_{4} (a q, 4 M)$ at $298 K$
Calculate the EMF of the cell.
Given:
$E_{A g^{+} / A g}^{0} = 0.80 V$
$E_{C u^{2 +} / C u}^{0} = 0.34 V$

E_{c e l l} = 0.301 V

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E_{c e l l} = 0.383 V

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E_{c e l l} = 0.45 V

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E_{c e l l} = 1.04 V

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Solution

The correct option is B $E_{c e l l} = 0.383 V$
Standard reduction potential of $A g^{+} / A g$ couple has value higher value, hence, it will act as cathode and $C u^{2 +} / C u$ will act as anode.

Thus, the cell representation will be,
$C u (s) ∣ C u^{2 +} (a q, 4.0 M) ∥ A g^{+} (a q, 0.1 M) ∣ A g (s)$
Cell reaction is
$C u + 2 A g^{+} \to C u^{2 +} + 2 A g$
Here $n = 2$ ;
$E = E_{c e l l}^{0} - \frac{0.059}{2} l o g \frac{[C u^{2 +}]}{[A g^{+}]^{2}}$
$E_{c e l l}^{0} = E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$
$E_{c e l l}^{0} = 0.80 - 0.34$
$E_{c e l l}^{0} = 0.46 V$

By Nernst equation,
$E_{c e l l} = 0.46 - \frac{0.059}{2} l o g \frac{[4.0]}{{[0.1]}^{2}}$
$E_{c e l l} = 0.46 - \frac{0.059}{2} l o g (4 \times 10^{2})$
$E_{c e l l} = 0.46 - \frac{0.059}{2} \times 0.602 - 0.059$
$E_{c e l l} = 0.46 - 0.018 - 0.059$
$E_{c e l l} = 0.46 - 0.077$
$E_{c e l l} = 0.383 V$

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Similar questions

The e.m.f. of the cell at $25^{\circ} C$
$\frac{C u}{C u^{2 +} (0.01 M)} | | \frac{A g^{+} (0.1 M)}{A g}$
is $[E^{0} (\frac{C u^{2 +}}{C u}) = 0.34 V a n d E^{0} (\frac{A g +}{A g}) = 0.80 V]$

Q. EMF of the cell,

A g | A g N O_{3} (0.1 M) | K B r (1 N), A g B r (s) | A g

- 0.6 V

298 K

Calculate the

K_{s p}

A g B r

298 K

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. Find the valency of copper from the following cell at

298 K

, if the emf of the cell is

0.0295 V

C u (s) | C u S O_{4} (a q, 0.1 M) | | C u S O_{4} (a q, 1 M) | C u (s)

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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Gibb's Free Energy

Standard XII Chemistry

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A galvanic cell is formed by immersing Ag(s) in AgNO3(aq,0.1 M) and Cu(s) in CuSO4(aq,4M) at 298 K Calculate the EMF of the cell. Given: E0Ag+/Ag=0.80 V E0Cu2+/Cu=0.34 V

A galvanic cell is formed by immersing $A g (s)$ in $A g N O_{3} (a q, 0.1 M)$ and $C u (s)$ in $C u S O_{4} (a q, 4 M)$ at $298 K$
Calculate the EMF of the cell.
Given:
$E_{A g^{+} / A g}^{0} = 0.80 V$
$E_{C u^{2 +} / C u}^{0} = 0.34 V$