The emf of the cell in which the following reaction- Zns - Ni2- 0-1 M - Zn2- 1-0 M - Nisoccurs- is found to be 0-5105 V at 298 K- The standard emf of the cell is-

The correct option is D 0.5400 V E _ cell = E _ 0 0.059 / n log Q ; E _ 0 =standard emf Zn(s)+ Ni ^ 2+ (0.1M) #8594; Zn ^ 2+ (1.0M)+N ...

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Standard Electrode Potentials

The emf of th...

Question

The emf of the cell in which the following reaction,

$Z n (s) + N i^{2 +} (0.1 M) \to Z n^{2 +} (1.0 M) + N i (s)$
occurs, is found to be $0.5105 V$ at $298 K$ . The standard emf of the cell is:

0.4810 V

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0.5696 V

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- 0.5105 V

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0.5400 V

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Z n (s) + N i^{2 +} (1 M) \to Z n^{2 +} (1 M) + N i (s)

0.5105 V .

Z n {/ Z n}^{2 +} ∥ {N i}^{2 +} / N

E^{o}

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

E^{o}

{N i}^{2 +} / N i = - 0.25 V

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

Z n^{2 +} | Z n

N i^{2 +} | N i

- 0.76 V

- 0.25 V

Z n + {C u}^{2 +} ⟶ C u + {Z n}^{2 +}

1.10 V

25^{o} C

0.1 M

{C u}^{2 +}

0.1 M

{Z n}^{2 +}

25^{o} C

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

E_{c e l l}^{o} = - 1.562 V

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