The standard electrode potentials of $\frac{Z n^{2 +}}{Z n}$ and $\frac{A g^{+}}{A g}$ are $- 0.763 V$ and $+ 0.799 V$ respectively. The standard potential of the cell is

1.562V

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2.361V

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– 1.562 V

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– 2.361V

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Solution

The correct option is A 1.562V
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ} = 0.799 - (- 0.763) = 1.562 V$

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The standard electrode potentials of Zn²⁺/Zn and Ag⁺/Ag are -0.763 V and +0.799 V respectively. The standard potential of the cell is

[KCET 1993]

The standard electrode potentials of $\frac{Z n^{2 +}}{Z n}$ and $\frac{A g^{+}}{A g}$ are $- 0.763 V$ and $+ 0.799 V$ respectively. The standard potential of the cell is

Q. The values of standard electrode potentials of

{Z n}^{2 +} / Z n

and

{A g}^{+} / A g

are

- 0.763 V

and

+ 0.799 V

respectively. The standard potential of the cell is:

Q. The standard electrode potentials of

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and

A g^{+}

| Ag are -0.763 V and +0.799 V respectively. The standard potential of the cell is

Q. The standard electrode potentials of

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and

C u

are

- 0.76, 0.80

and

0.34

volt respectively then_________.

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The standard electrode potentials of Zn2+Zn and Ag+Ag are −0.763 V and +0.799 V respectively. The standard potential of the cell is

The correct option is A 1.562VE∘cell=E∘cathode−E∘anode=0.799−(−0.763)=1.562 V

The standard electrode potentials of $\frac{Z n^{2 +}}{Z n}$ and $\frac{A g^{+}}{A g}$ are $- 0.763 V$ and $+ 0.799 V$ respectively. The standard potential of the cell is

The correct option is A 1.562V
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ} = 0.799 - (- 0.763) = 1.562 V$