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EMF

If Ag⊕ | Ag...

Question

If $A g^{\oplus} | A g$ half cell in the above voltaic cell is replaced by $Z n^{2 +} | Z n$ half cell:
$(E_{Z n^{2 +} | Z n}^{⊖} = - 0.76, V)$

The direction of current flow will remain same.

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Polarity of the electrodes will be reversed.

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Cell will stop working.

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EMF of the cell will increase.

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Solution

The correct option is B Polarity of the electrodes will be reversed.
As cell operates, $[X^{2 +}]$ will increase and $[A g^{\oplus}]$ will decrease.
[Cell reaction : $X + 2 A g^{\oplus}, \to 2 A g + X^{2 +}$ ]
If $Z n^{2 +} | Z n$ replaces $A g^{\oplus} | A g$ , cell polarities will change.
$E_{c e l l}^{⊖} = - 0.25 - (- 0.76) = 0.51 V$

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Similar questions

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

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

E^{o}

{Z n}^{2 +} | Z n

- 0.76

Z n ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ ∣ ∣ 2 H^{+} (1 M) ∣ ∣ H_{2} (1 a t m)

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

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

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?

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

E_{c e l l}

Z n | Z n^{2 +} (0.1 M) | | A g^{+} (1.0 M | A g

298 K

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