The standard hydrogen electrode has a zero electrode potential because .....

  • A
    Hydrogen is easily oxidized.
  • B
    The electrode potential is assumed to be zero.
  • C
    The hydrogen atom has only one electron.
  • D
    Hydrogen is the lightest element.

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Similar Questions

The standard reduction potential at $298 \ K$ for the following half-cell reactions is given as:
$Zn^{2+}_{(aq)} + 2e^{-} \rightarrow Zn_{(s)} ; \quad E^{\circ} = -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightarrow Cr_{(s)} ; \quad E^{\circ} = -0.740 \ V$
$2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)} ; \quad E^{\circ} = 0.0 \ V$
$F_{2(g)} + 2e^{-} \rightarrow 2F^{-}_{(aq)} ; \quad E^{\circ} = 2.87 \ V$
Which of the following is the strongest reducing agent?

The standard electrode potentials for $Ag^{+}/Ag$,$Hg_2^{2+}/2Hg$,$Cu^{2+}/Cu$,and $Mg^{2+}/Mg$ are $0.80 \, V$,$0.79 \, V$,$0.34 \, V$,and $-2.37 \, V$ respectively. If an aqueous solution containing $1 \, M$ concentration of each of these metal ions is electrolyzed,what is the correct order of metal deposition at the cathode as the voltage is increased?

Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Given at $298 \ K$: $E^\ominus_{Fe^{2+}/Fe} = X \ V$; $E^\ominus_{Fe^{3+}/Fe} = Y \ V$. The $E^\ominus_{Fe^{3+}/Fe^{2+}}$ in Volt at $298 \ K$ is given by:

Calculate the $emf$ of the cell $Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^+_{(aq)} | Ag_{(s)}$. Given: $E^0_{Cu^{2+}/Cu} = +0.34 \ V$,$E^0_{Ag^+/Ag} = +0.80 \ V$.

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