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Nernst Equation

Calculate the...

Question

Calculate the EMF of a Daniel cell when the concentration of $Z n S O_{4}$ and $C u S O_{4}$ are $0.001$ M and $0.1$ M respectively. The standard potential of the cell is $1.1$ V.

A

E = 1.159 V

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B

E = 2.159 V

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C

E = 1.870 V

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D

None of these

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Solution

The correct option is A $E = 1.159 V$
According to nernst equation,,
$E = 1.1 + 2.303 R T / n F l o g [Z n^{+ 2}] / [C u^{+ 2}] = 1.1 + 2.303 R T / 2 F l o g 0.001 / 0.1 = 1.159 V$

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Similar questions

Q. Calculate the EMF of a Danial cell when the concentration of

Z n S O_{4}

C u S O_{4}

0.001

0.1

M respectively. The standard EMF of the cell is

1.1

Q. The emf of a Daniel cell at 298 K is

E_{1}

concentration of

Z n S O_{4}

is 0.1 M and that

C u S O_{4}

is 0.01 M, the emf changed to

E_{2}

.What is the relationship between

E_{1}

E_{2}

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 +}

Q. The emf of Daniell cell at

298 K

E_{1} Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) | C u

When the concentration of

Z n S O_{4}

1.0 M

C u S O_{4}

0.01 M

, the emf changed to

E_{2}

What is the relation between

E_{1}

E_{2}

?

V

) of a Daniel cell containing

0.1 M Z n S O_{4}

0.01 M C u S O_{4}

solutions at their respective electrodes are:

(E_{{C u}^{2 +} / C u}^{o} = 0.34 V; E_{{Z n}^{2 -} / Z n}^{o} = - 0.76 V)

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