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Standard XII

Chemistry

Electrode Potential

The emf of th...

Question

The emf of the cell involving the following reaction, ${2 A g}^{+} + H_{2} ⟶ 2 A g + {2 H}^{+}$ is $0.80$ volt. The standard oxidation potential of silver electrode is:

- 0.80

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0.50

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- 0.30

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0.40

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Solution

The correct option is A $- 0.80$ V
$E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}$
$0.80 = E_{{A g}^{+} / A g}^{o} + E_{H_{2} / H^{+}}^{o}$
$0.80 = E_{{A g}^{+} / A g}^{o} - 0$
$E_{o x i d a t i o n}^{o} = - E_{r e d}^{o} = - 0.80 V$

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Similar questions

Q. The standard e.m.f. of the cell involving the reaction,

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Q. Consider the reaction,

2 A g^{+} + C d \to 2 A g + C d^{2 +}

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The emf of the cell involving the following reaction, 2Ag++H2⟶2Ag+2H+is 0.80 volt. The standard oxidation potential of silver electrode is:

The correct option is A −0.80 VEocell=Eocathode−Eanode0.80=EoAg+/Ag+EoH2/H+0.80=EoAg+/Ag−0Eooxidation=−Eored=−0.80V

The emf of the cell involving the following reaction, ${2 A g}^{+} + H_{2} ⟶ 2 A g + {2 H}^{+}$ is $0.80$ volt. The standard oxidation potential of silver electrode is:

The correct option is A $- 0.80$ V
$E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}$
$0.80 = E_{{A g}^{+} / A g}^{o} + E_{H_{2} / H^{+}}^{o}$
$0.80 = E_{{A g}^{+} / A g}^{o} - 0$
$E_{o x i d a t i o n}^{o} = - E_{r e d}^{o} = - 0.80 V$