$A$ standard hydrogen electrode has zero electrode potential because

  • A
    Hydrogen is easiest to oxidise
  • B
    The electrode potential is assumed to be zero
  • C
    Hydrogen atom has only one electron
  • D
    Hydrogen is the lightest element

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Using the data given below,find out the strongest reducing agent:
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$

If the $emf$ of the cell $Cu(s) | Cu^{2+}(1 \text{ M}) || Ag^+(1 \text{ M}) | Ag(s)$ is $0.463 \text{ V}$ at $25^{\circ} \text{C}$ and the standard electrode potential of the $Cu$ electrode is $0.337 \text{ V}$, find the standard electrode potential of the $Ag$ electrode. (in $\text{ V}$)

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

What is electrode potential in a galvanic cell? Explain in detail.

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